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Grades 5-10 Young Mathematical Circle Plan and Developments
QUVA DISTRICT
OWNED BY XTMFMTTEB
4 SECONDARY SCHOOL
MATHEMATICS TEACHER
ERGASHOV JALOLIDDIN
"YOUNG MATHEMATICS"
CIRCLE
DOCUMENTS
2016-2017 academic year
"I approve"
School principal: D.Eraliyeva
“___” _____________2017
"Young mathematician"
annual work plan of the circle.
| number | Topics | Source | Soat | Calendar time | Transition time |
| 1 | Muhammad ibn Musa al-Khwarizmi is the great mathematician of the world. | Mathematics on stage | 1 | ||
| 2 | Signs of division of numbers. | Mathematics on stage | 1 | ||
| 3 | Linear function and its graph | Algebra-8 | 1 | ||
| 4 | Mathematical focus: "Wonderful memory". | Mathematics on stage | 1 | ||
| 5 | System of linear equations. | Algebra-8 | 1 | ||
| 6 | Ghiyosiddin Jamshid Kashi. | Mathematics on stage | 1 | ||
| 7 | Methods for solving systems of equations. | Algebra-8 | 1 | ||
| 8 | Write numbers with action symbols and the same numbers. | Don't count to eight | 1 | ||
| 9 | Solve problems using a system of equations. | Algebra-8 | 1 | ||
| 10 | Roman numerals. | Mathematics on stage | 1 | ||
| 11 | Numerical inequalities and their properties. | Algebra-8 | 1 | ||
| 12 | Mathematical game "GREAT". | Mathematics on stage | 1 | ||
| 13 | Addition and multiplication of inequalities | Algebra-8 | 1 | ||
| 14 | Mathematical sophism. | Mathematics on stage | 1 | ||
| 15 | Solve an unknown inequality. | Algebra-8 | 1 | ||
| 16 | Solving systems of inequalities. | Algebra-8 | 1 | ||
| 17 | ECUB. | Mathematics-6 | 1 | ||
| 18 | ECUK. | Mathematics-6 | 1 | ||
| 19 | Find the two numbers by their sum and ratio. | Problem solving | 1 | ||
| 20 | Find the two numbers by their difference and ratio. | Problem solving | 1 | ||
| 21 | Find two numbers using their sum and subtraction. | Problem solving | 1 | ||
| 22 | Speed detection issues. | Problem solving | 1 | ||
| 23 | Meeting action issues. | Problem solving | 1 | ||
| 24 | Chase Actions. | Problem solving | 1 | ||
| 25 | Replacing one quantity with another. | Problem solving | 1 | ||
| 26 | e number date. | History of mathematics | 1 | ||
| 27 | Equalize the data and subtract one from this. | Problem solving | 1 | ||
| 28 | Collaborative work. | Problem solving | 1 | ||
| 29 | Find the two multipliers, their given multipliers and their multipliers, when they are equal. | Problem solving | 1 | ||
| 30 | Problems to be solved from the end. | Problem solving | 1 | ||
| 31 | Interesting and issues related to different life situations. | Problem solving | 1 | ||
| 32 | History of the number Pi. | History of mathematics | 1 | ||
| 33 | Problems that can be solved by assumption. | Problem solving | 1 | ||
| 34 | Math night. | Event | 1 |
MMIBDO ': / / B.Teshaboyev
Date: ______
TOPIC 1: Muhammad ibn Musa al-Khwarizmi, the great mathematician of the world.
Muhammad ibn Musa al-Khwarizmi was born in 787 in ancient Khorezm. Although Al-Khwarizmi was ten years old, his brain seemed busy, and his brain was busy thinking of hundreds of solutions to complex problems and examples. However, as the situation in his homeland became more and more difficult, al-Khwarizmi left Khorezm and went to Babylon. In Baghdad, the capital of the Khilafah, Muhammad ibn Musa al-Khwarizmi wrote the famous royal work, The Elephant, al-Hind, who had his own independent opinion. comes as a renowned, talented young scientist. Harun ar-Rashid al-Khwarizmi greeted him with sweet words, honor and invited him to work in his palace. Harun al-Rashid gathered the famous scholars of that time in Baghdad and entrusted al-Khwarizmi to lead them.
Knowing that the scholar was a man of strong thought and knowledge, Harun al-Rashid al-Khwarizmi fearlessly endorsed the daring idea of organizing the House of Wisdom in Baghdad and financially supported the House of Science. The caliph Harun al-Rashid died suddenly in 807, when al-Khwarizmi was in charge of the construction and was busy putting it into operation as soon as possible. After his death, his son al-Ma'mun ascended the throne. The caliphate of al-Ma'mun coincided with the heyday of al-Khwarizmi's scientific activity.
At the suggestion of al-Khwarizmi, the great mathematicians and famous astronomers of the time, such as Muhammad al-Farghani, Ahmad al-Murwazi, Abbas al-Gawhari, Tahir Yassavi, Riza Turkistani, moved from Turkestan to Baghdad, and world science created a miracle of development in the history of which was later called the "Arab school of mathematics."
Al-Khwarizmi and his countrymen made universal discoveries, and the ancient Greek scientist Erotosthenes clarified the calculations on the Sanjar plateau and measured the length of one degree of the Earth's meridian. This dimension later played an important role in the development of astronomy and geography.
The Baghdad Mathematical School "Baytul-Hikma" under the leadership of Al-Khwarizmi has left an indelible mark on the history of world culture. Mamun's table of astronomy, his book of pictures of the universe, and a number of his great works in the fields of mathematics and astronomy, geography, and geodesy play an important role in the development of these sciences in the following centuries. The great scholar al-Khwarizmi, who left his home "until the end of the uprising," lived in Baghdad for forty-five years, devoting himself to science and even his family. He died at the age of 63, without having children.
MMIBDO ': / / B.Teshaboev
Date: ______
TOPIC 2:Signs of division of numbers.
- 2 signs of division.
If the last digit of a given number is an even number, or zero, that number itself is also divisible by 2 without a remainder.
- 3 signs of division.
If the sum of the digits of a given number is divisible by 3, then that number itself is also divisible by 3 without a rule.
- Signs of division into 4.
A number consisting of the last two digits of a given number is divisible by 4, or if the last two digits are 0, the given number is divisible by 4.
- 5 signs of division.
Numbers ending in 0 or 5 are divisible by 5 without remainder.
- 6 signs of division.
If a given number is divisible by 2 and 3, these numbers are divisible by 6 without remainder.
- 7 signs of division.
If the given number is divided by 7 and the difference is divided by 7, the given number is divided by XNUMX.
- 8 signs of division.
If the last three digits of a given number are divisible by 0 or 8, the given number is divisible by 8.
- 9 signs of division.
Numbers whose sum of numbers is divisible by 9 are divisible by 9 without remainder.
- 10 signs of division.
Numbers with the last digit 0 are divisible by 10.
- 25 signs of division.
If the last two digits are divisible by 0 or 25, the given number is divisible by 25.
MMIBDO ': / / B.Teshaboev
Date: ______
TOPIC 3: Linear function and its graph.
A linear function is a function of the form y = kx + b, where k and b are given numbers. When b = 0, the linear function has the form y = kx, and its graph is a straight line passing through the origin. Based on this fact, it can be shown that the graph of the linear function y = kx + b is a straight line. Since only one straight line passes through two points, it is enough to make two points of this graph to make a graph of the function y = kx + b.
Issue 1. Graph the function y = 2x + 5.
x When = 0, y = 2x The value of the function + 5 is equal to 5, i.e. (0; 5) belongs to the point graph.
Agar x If = 1, then y = 2 · 1 + 5 = 7, ie the point (1; 7) also belongs to the graph. Draw points (0; 5) and (1; 7) and draw a straight line through them. This is a straight line y = 2x + 5 is the graph of the function ▲
y = 2x The + 5 function graph is the ordinate of each point y = 2x we see that the graph of the function is 5 units larger than the ordinate of that abscissa. This is it y = 2x + 5 Each point of the function graph y=2x means that the corresponding point on the graph of the function is formed by moving 5 units up along the ordinate axis.
In general, the graph of the function y = kx + b is formed by moving the graph of the function y = kx along the ordinate axis to the unit b. The graphs of the functions y = kx and y = kx + b are parallel straight lines
Issue 2. y =-2x Find the points of intersection of the graph of the function + 4 with the coordinate axes.
Find the point of intersection of the graph with the abscissa axis. The ordinate of this point is 0. Therefore -2x + 4 = 0, hence x = 2.
Thus, the point of intersection of the graph with the abscissa axis has a coordinate (2; 0).
Find the point of intersection of the graph with the ordinate axis. Since the abscissa of this point is 0 y = -2 · 0 + 4 = 4.
Thus, the point of intersection of the graph with the ordinate axis has a coordinate (0; 4) (Figure 16).
Exercises
- 1) There were 400 tons of potatoes in the vegetable warehouse. Every day another 50 tons of potatoes were delivered to the warehouse. Quantity of potatoes (p) of time (t) with the formula.
- The tourist left the city by bus for 10 km, and then began to walk in the same direction at a speed of 5 km / h. Sayyoh x how many hours after walking from the city (y) was at a distance ?.
MMIBDO ': / /
TOPIC 4: Mathematical focus: "Wonderful memory".
While performing this trick, the student goes to the members of the circle and says to them, “I want to show you how wonderful my memory is. The rectangular papers in my hand have the serial number and the seven-digit number written on them. I will distribute these papers to you. You take turns saying the serial number of this paper, and I immediately count and tell you the seven-digit number written on it. ” So the magician distributes the rectangular pieces of paper to the members of the circle. They take turns raising their hands and saying the different ordinal numbers on the paper, and the magician keeps writing the seven-digit number on it on the board. For example, if a student says 13, the magician writes and reads 4 million 718 thousand 976 on the board. After this is repeated several times, the magician asks the students: - Tell me, did I remember these numbers, or is there a secret in it?
The numbers on the paper distributed by the magician are formed according to different laws.
Method 1 For example, let the sequence number on the paper be a two-digit number, that is, take the following view:
| № 23
5831459 |
The formation of the seven-digit number written on this rectangular paper is as follows: the sum of the ordinal numbers 2 and 3 is 2 + 3 = 5; the sum of the next number 3 and 5 is 3 + 5 = 8; The sum of 5 and 8 is 5 + 8 = 13 (where the last number is 3); 8 + 3 = 11 (the last number is written as 1), etc. Seven numbers are generated. If the serial number on the special paper is a one-digit number, that is, if:
| № 2
4606628 |
In this case, 2 is added to itself to form 4, and the remaining numbers are formed as above. 2 + 4 = 6; 4 + 6 = 10 (0 is written) and so on.
MMIBDO ': / / B.Teshaboev
Date: ______
TOPIC 5: Methods for solving systems of equations.
- The replacement method is as follows:
1) from one of the equations of the system (no matter which one) it is necessary to express one unknown by another, for example, y by x;
2) the resulting expression must be put in the second equation of the system, an unknown equation is formed;
3) solve this equation and find the value of x;
4) Find the value of y by putting the found value of x in the expression for y
Solve the system of equations:
In the system of equations we change the form (to the common denominator):
1) 9x+2y= 12, 2y= 12-9x,
2)
3)
Answer: x= 0, y= 6. ▲
- To solve a system of equations by the method of algebraic addition:
1) equalization of the modulus of the coefficients in front of one of the unknowns;
2) to find one unknown by adding or subtracting the formed equations;
3) Put the found value in one of the equations of the given system and find the second unknown.
Solve the system of equations.
| (2) |
1) Leaving the first equation unchanged, multiply the second equation by 4:
| (3) |
2) Subtracting the first equation from the second equation of the system (3), we find: 11y = -22, hence y = -2.
3) y Substituting = -2 into the second equation of the system (2), we find: x + 2 · (-2) = -2, hence x = 2.
Answer: x = 2, y = -2. ▲
- The graphical method of solving a system of equations is as follows:
1) a graph of each equation of the system is made;
2) find the coordinates of the point of intersection of the straight lines (if they intersect). The coordinates of the point of intersection of the graphs of the equations are the solution of this system of equations.
There can be three cases in the arrangement of two straight lines in the plane - the graphs of the system of equations:
1) straight lines intersect, ie have a common point. In this case, the system of equations has a single solution
2) straight lines are parallel, ie they do not have common points. In this case, the system of equations has no solutions;
3) straight lines overlap. In this case, the system has an infinite number of solutions.
Issue 1. Show that the following system of equations has no solutions:
Multiply the first equation of the system by 2 and subtract the second equation of the given system from the resulting equation:
_ 2x + 4y = 12
2x + 4y = 8
_______________________
= 0 4
Incorrect equality. So, x va y (5) does not have values from which both equations of the system can be true, that is, (5) the system has no solutions. ▲
This means that, from a geometric point of view, the graphs of system equations (5) are parallel straight lines. (Figure 20)
- Solve the system of equations by substitution:
1) 2) 3)
- Solve the system of equations by algebraic addition:
1) 2) 3)
- Solve the system of equations graphically:
1) 2) 3)
MMIBDO ': / / B.Teshaboyev
Date_____
TOPIC 6: Ghiyosiddin Jamshid Kashi.
One of the great scientists of Ulugbek scientific school is Jamshid Koshi. Kashi was born in 1385 in the city of Kashan. From a young age, Cauchy became famous as a leading mathematician and astronomer of his time. Ulugbek also invited him to Samarkand, and Kashi came to Samarkand in 1417 and took an active part in the construction of the Ulugbek observatory, carrying out great scientific work.
He described the results of his scientific work in 10 works on astronomy and 3 works on mathematics. One of Jamshid Kashi's works is The Key to Arithmetic. This work is an encyclopedia of medieval elementary mathematics. Cauchy wrote this work in 1427. The Key to Arithmetic consists of an introduction and five parts. The introduction consists of 6 chapters, which describe the definition of arithmetic, number and its types.
The second part is devoted to the arithmetic of fractions and consists of 12 chapters. In this section, he described important ideas about different fractions, operations on them, and decimal fractions. Cauchy introduced the terms of reading and writing fractions with denominators of 10, 100, 1000,…, that is, decimal fractions. Cauchy describes these fractions and explains how to read "ten", "hundred", "thousand",…, and when writing, write the fractional part after the whole part, or write the whole part of the decimal fraction in a different color ink. Gives many examples of operations on decimal fractions. Thus, Cauchy was the first scientist to establish the theory of decimal fractions.
In 1424 in Samarkand the highest peak of development of Kashi's work "The treatise on a circle" is considered. It is known that the ratio of the length of a circle to its diameter is a constant number, denoted by the letter "". In this play, Cauchy determines the value of "" with a very high precision with 17 digits after the comma.
“” = 3,14159265358997932.
Cauchy’s calculations seen above are of great accuracy, amaze everyone, and Cauchy leaves an indelible mark on the history of mathematics.
MMIBDO ': / / B.Teshaboev
Date: ______
TOPIC 7: Methods for solving systems of equations.
- The replacement method is as follows:
1) from one of the equations of the system (no matter which one) it is necessary to express one unknown by another, for example, y by x;
2) the resulting expression must be put in the second equation of the system, an unknown equation is formed;
3) solve this equation and find the value of x;
4) Find the value of y by putting the found value of x in the expression for y
Solve the system of equations:
In the system of equations we change the form (to the common denominator):
1) 9x+2y= 12, 2y= 12-9x,
2)
3)
Answer: x= 0, y= 6. ▲
- To solve a system of equations by the method of algebraic addition:
1) equalization of the modulus of the coefficients in front of one of the unknowns;
2) to find one unknown by adding or subtracting the formed equations;
3) Put the found value in one of the equations of the given system and find the second unknown.
Solve the system of equations.
| (2) |
1) Leaving the first equation unchanged, multiply the second equation by 4:
| (3) |
2) Subtracting the first equation from the second equation of the system (3), we find: 11y = -22, hence y = -2.
3) y Substituting = -2 into the second equation of the system (2), we find: x + 2 · (-2) = -2, hence x = 2.
Answer: x = 2, y = -2. ▲
- The graphical method of solving a system of equations is as follows:
1) a graph of each equation of the system is made;
2) find the coordinates of the point of intersection of the straight lines (if they intersect). The coordinates of the point of intersection of the graphs of the equations are the solution of this system of equations.
There can be three cases in the arrangement of two straight lines in the plane - the graphs of the system of equations:
1) straight lines intersect, ie have a common point. In this case, the system of equations has a single solution
2) straight lines are parallel, ie they do not have common points. In this case, the system of equations has no solutions;
3) straight lines overlap. In this case, the system has an infinite number of solutions.
Issue 1. Show that the following system of equations has no solutions:
Multiply the first equation of the system by 2 and subtract the second equation of the given system from the resulting equation:
_ 2x + 4y = 12
2x + 4y = 8
_______________________
= 0 4
Incorrect equality. So, x va y (5) does not have values from which both equations of the system can be true, that is, (5) the system has no solutions. ▲
This means that, from a geometric point of view, the graphs of system equations (5) are parallel straight lines. (Figure 20)
- Solve the system of equations by substitution:
1) 2) 3)
- Solve the system of equations by algebraic addition:
1) 2) 3)
- Solve the system of equations graphically:
1) 2) 3)
MMIBDO ': / / B.Teshaboyev
Date_____
TOPIC 8: Writing numbers with action symbols and the same numbers.
- Write the number 3 with five 37 digits and action signs.
3 + 3 3 + 3: 3 = 37
- Write the number 2 using four 111 numbers and action signs.
2 2 2: 2 = 111
- Write the number one thousand using the number five 9 and the action symbols.
9: 9 + 9 9 9 = 1000
- Create the number 2 using 28 XNUMX and just the addition operation.
2 2 + 2 + 2 +2 = 28
- How do you write 101 with six identical numbers?
aaaa: aa = 101
- Write the number 1, provided that the numbers are in ascending order, using numbers from 9 to 100 and operation symbols.
1 + 2 + 3 + 4 + 5 + 6 + 7 + (8 * 9) = 100
1 + 2 + (2 * 3) + (4 + 5) + 6 - 7 + 8 * 9 = 100
1 * 2 + 3 4 + 5 6 + 7 - 8 + 9 = 100
- Write the number 30 using three identical numbers and operations.
6 * 6 - 6 = 30 33+ 3 = 30
5 * 5 + 5 = 30 3 3 - 3 = 30
- Write the number million using only 3 numbers and operations.
((333-33): 3)3= 1000000
- Write 24 using three double numbers and actions.
2 2 + 2 = 24
- Write the numbers 2 to 20 with the number 25 in five.
2 2 - 2 - 2 + 2 = 20 2 2 - 2 + (2: 2) = 21
2 2 * 2 - 2 2 = 22 2 2 + 2 - (2: 2) = 23
2 2 - 2 + 2 + 2 = 24 2 2 + 2 + (2: 2) = 25
MMIBDO ': / / B.Teshaboev
Date: ______
Topic 9: Solving problems using a system of equations.
Problem solving using a system of equations is usually carried out according to the following scheme:
1) definitions for the unknown are made and a system of equations corresponding to the content of the problem is created;
2) the system of equations is solved;
3) Return to the condition of the case and write the answer.
Masala. If the sum of two numbers is more than 5 times their difference, and the sum of these numbers is more than 8 times their difference, find these numbers.
1) Create a system of equations.
Let's say x, y - be the numbers sought. In this case, depending on the condition of the problem, we have:
(3)
2) Solve the system.
First we simplify the equations of the system (3):
(4)
Divide the second equation in (4) by 2 and divide it by the first equation: _ x + 3y = 5
x + 2y = 4
___________
y = 1
y Substituting = 1 (4) into the first equation of the system, x + 3 · 1 = 5, x We find that = 2.
Answer. The numbers sought are 2 and 1. ▲
Issue 1:
|
|
13000 kg of sugar and 4 kg of high-grade flour were purchased for 7 soums. If 3 kg of flour costs 1300 soums more than two kilograms of sugar, find the price of 1 kg of sugar and 1 kg of flour. |
| A | 1150 soums, 1250 soums |
| B | 1150 soums, 1200 soums |
| C | 100 soums, 1350 soums |
| D | 1200 soums, 1100 soums |
Issue 2
| The first student worked for 3 hours and the second for 2 hours and made 36 details together. If they made 1 parts together in 14 hour, how many parts did each of them make? | |
| A | 24, 12 |
| B | 30, 16 |
| C | 18, 18 |
| D | 14, and 22 ta |
Issue 3
| Two types of 2000 kg of biscuits were purchased for the kindergarten for 2500 soums and 10 soums, respectively, and 22000 soums were paid for all of them. How many kilograms are obtained from each type of biscuit? | |
| A | 6 kg, 3 kg |
| B | 5 kg, 5 kg |
| C | 6 kg, 4 kg |
| D | 3 kg, 7 kg |
Issue 4
| 4 kg of fodder per day was allocated for 10 horses and 88 cows. If it is known that 2 horses were given 5 kg more than 4 cows, how much feed was given to each horse and each cow per day? | |
| A | 12 kg, 3 kg |
| B | 10 kg, 6 kg |
| C | 12 kg, 4 kg |
| D | 12 kg, 6 kg |
MMIBDO ': / / B.Teshaboyev
Date_____
TOPIC 10: Roman numerals.
Roman numerals should be known to every civilized person, as they are still used in writing dates, making lists, marking chapters and sections in books, and so on. Students are shown the following table and explained the Roman numerals and their values in the decimal number system.
| Roman numerals | I | V | X | L | C | D | M |
| Their value | 1 | 5 | 10 | 50 | 100 | 500 | 1000 |
The origin of Roman numerals is directly related to the names of letters in the Latin alphabet: I - ,, i ”; V - ,, ve ”; X - ,, iks ”; L - ,, el ”; C - ,, se ”; D - ,, de ”; M - ,, em ”; any number up to a million is written using these letters. There are certain rules when writing numbers with Roman numerals, i.e. a single number cannot be written side by side more than three times when writing a number.
Writing order: I-one; II-two; III-uch; IV-four; V-besh; VI-six; VII-yetti; VIII-eight; IX-nine; X-on. Similar numbers from 20 to XX can be written in the same way: XI; XII; XIII; XIV; XV; XVI; XVII; XVIII; XIX; XX; ……
When determining the value of numbers written in Roman numerals, care should be taken that if a small number is written to the left of a large number, the number of units in the small number is subtracted from the number of units in the large number. If a small number is written to the right of a large number, the number of units in the small number is added to the number of units in the large number.
1-misol. XXXVII=10+10+10+5+1+1=37 CLXIII=100+50+10+1+1+1=163 CXL=100+(50- 10)=140 XL=50-10=40
2-misol. 102=100+2=CII 374=100+100+100+50+10+10+(5-10)=CCCLXXIV
Large numbers such as 29635 are written as follows:
XXIXmDCXXXV = (10 + 10 + (10-1)) m + 500 + 100 + 10 + 10 + 10 + 5 The lowercase m is derived from the Latin word mille, which denotes the number thousand.
Exercises:
- Write the following numbers in Arabic numerals: XXIII, XXXIV, DXIV, MDCLXVI, DmIX, MCXLVI, XXXIV, XXIX, CDXXI, CMIII, MCMXLV.
- Express these numbers in Roman numerals: 49, 574, 1147, 1974, 5003.
MMIBDO ': / / B.Teshaboev
Sana: ______
Topic 11: Numerical inequalities and their properties.
If a> b and b> c, then a> c.
If the same number is added to both parts of the inequality, then the sign of the inequality does not change.
Any join can be moved from one part of the inequality to another by substituting the sign of that join for the opposite.
If both parts of the inequality are multiplied by the same positive number, then the sign of the inequality does not change.
If both parts of the inequality are multiplied by the same negative number, then the sign of the inequality changes to the opposite.
If both parts of the inequality are divided by the same positive number, then the sign of the inequality does not change. If both parts of the inequality are divided by the same negative number, then the sign of the inequality changes to the opposite
Issue 1. Agar a>b if so -a<-b Prove that
>b multiplying both parts of the inequality by -1 negative number, -a<-b we create. ▲
For example, the inequality 1,9 <2,01 gives rise to the inequality -1,9> -2,01, and the inequality leads to the inequality.
Issue 2. Agar a va b - positive numbers and a>b If so, prove that it is.
b <a both parts of the inequality ab Divide by a positive number to form ▲
Issue 1
| If so, which of the following inequalities is valid? | |
| A | |
| B | |
| C | |
| D |
Issue 2
| If we divide both sides of the given inequality, what inequality is formed? | |
| A | |
| B | |
| C | |
| D |
Issue 3
| What inequality is formed if the inequality is divided into two parts? | |
| A | |
| B | |
| C | |
| D |
Issue 4
| What inequality is formed by multiplying both parts of a given inequality by? | |
MMIBDO ': / / B.Teshaboyev
Date_____
TOPIC 12: Mathematical game "GREAT".
Objective: To teach students to think fast, to be alert, to calculate mathematical operations verbally quickly and accurately.
This game teaches students to multiply and divide verbally quickly, helps to stabilize their attention, alertness, strengthen memory. In addition, students are very interested in this game and never get bored and play it over and over again. The game mode is as follows. Several students dial and count the numbers in order, starting at once. For example, when there are numbers that are divisible by 7 without a remainder and end in 7, the student who says that number must say the word "excellent" instead of that number. If the student does not say the word immediately or goes astray, the game stops, that student leaves the game, and the game begins anew, followed by the student. The only student to reach the end is the winner.
For example: Numbers divisible by 6:
1, 2, 3, 4, 5, "excellent", 7, 8, 9, 10, 11, "excellent", 13, 14, 15, "excellent", 17, "excellent", 19, 20, 21, 22, 23, “excellent”, 25, “excellent”, 27, 28, 29, “excellent” …… The game continues in this way.
MMIBDO ': / / B.Teshaboev
Date____
Topic 13: Addition and multiplication of inequalities
Theorem 1. Adding inequalities with the same sign gives the same sign inequality: if a> b and c> d, then a + c> b + d.
Examples:
1) 2)
Theorem 2. Multiplying inequalities of the same sign with positive left and right sides results in the same sign inequality: if a> b, c> d and a, b, c, d are positive numbers, in which case ac> bd.
Examples:
1) 2)
Agar a, b - positive numbers and a>b if so a2>b2 will be.
a> b Multiplying the inequality by itself, we get: a2>b2.
Similarly, a, b - positive numbers and a>b then any natural n for an>bn can be proved.
For example, 5> 3 from the inequality 55>35, 57>37 inequalities such as
Question 1
| Add the inequalities: and. | |
| A | |
| B | |
| C | |
| D |
Question 2
| Multiply the inequalities: and. | |
| A | |
| B | |
| C | |
| D |
Question 3
| Multiply the inequalities: and. | |
| A | |
| B | |
| C | |
| D |
Question 4
| If so, which of the following inequalities is correct? | |
| A | |
| B | |
| C | |
| D |
Question 6
| If and, then which of the following inequalities is valid? | |
| A | |
| B | |
| C | |
| D |
MMIBDO ': / / B.Teshaboyev
Date_____
Topic 14: Mathematical sophisms
There is a saying among the people that “twice is known as two,” which means that the assertion is logically proved on the basis of mathematical laws and the truths on which they are based. Therefore, if we take a logical contradiction on the basis of some reasoning, for example, the result of 2 × 2 = 5, it indicates that somewhere in our reasoning an error was made. But in many cases, this error is not easy to find.
In fact, at first glance, it is difficult to find fault with absolutely correct considerations:
- in that case and. By adding the last equations to the hadma had, we get the following, now we subtract or take ni from both sides. It follows from this.
2. We get the correct number equation: 225: 25 + 75 + 100-16 and after a few substitutions we get:
25(9:1+3)=84, 25×12=7×12, 5×5=7
3. We replace the equation as follows:
5005-2002=35×143-143×14
4.81-171 = 100-190 plus both sides of the equation
81-171 + = 100-190 +
or
;
in that case.
There is no proof of hyech here, only that the laws and rules of mathematics are violated. In the first example, the impossible operation is performed by dividing by zero (), and in the second, the distribution law of multiplication is incorrectly applied to the division operation ()
In the third case, the division by 0 is performed, and in the fourth case, the equality of the squares of the numbers leads to their equality (although equal).
Examples are given mathematical sophisms called Sophism (from the Greek word -puzzle, cunning) consists of a series of opinions close to the truth in which the error is hidden, thus leading to an absurd, paradoxical, contradictory conclusion;
Sophists have played a major role in the history of mathematics. They were the impetus for the discovery of new laws and the creation of theories. Sophisms are said to be solved if a mistake is found and thought out. The first book on sophisms was "Where's the Error?" By W. Litzman and F. Trier. his book was published in Petrograd in 1919, in which a number of mathematical sophisms were quoted and discussed.
MMIBDO ': / / B.Teshaboyev
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Topic 15: Solving an unknown inequality.
To solve an unknown inequality that leads to a linear inequality:
1) transfer of unknown participles to the left, and unknown participles to the right (property 1);
2) Summarize similar terms and divide both parts of the inequality by the coefficient in front of the unknown (if it is not equal to zero) (property 2).
Issue 1. Solve the inequality:
3(x-2)-4(x+1)<2(x-3)-2
Let's simplify the left and right parts of the inequality. We open the parentheses:
3x-6-4x-4<2x -6-2
We move the unknowns to the left of the inequality, and the unknowns (free) to the right (property 1):
3x-4x-2x<6 + 4-6-2
Summarize similar terms: - 3x <2
and divide both parts of the inequality by -3 (property 2):
Answer. ▲
This solution can be summarized as follows:
1) a At what values of is the fraction greater than the fraction?
2) b At what values of is the fraction smaller than the fraction?
3) x At what values of is the fraction greater than the difference of the fractions?
4) x At what values of and is the sum of the fractions smaller than the fraction?
Solve the inequality
1)
2)
3)
4)
5)
6)
7)
8)
MMIBDO ': / / B.Teshaboyev
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Topic 16: Solving systems of inequalities.
Issue 1. Solve the system of inequalities:
| (1) |
Solve the first inequality:
Thus, the first inequality xExecuted when> 2.
Solve the second inequality:
Thus, (1) is the second inequality of the system xExecuted when> -3.
On the number axis (1) we describe the sets of solutions of the first and second inequalities of the system.
Solutions of the first inequality x> 2 are all points of light, solutions of the second inequality x> -3 are all points of light
(1) system solutions x have values corresponding to both rays at the same time. As you can see from the picture, this is a set of all the common points of the rays xThere will be> 2 lights.
Answer. x> 2. ▲
Find all integers with solutions of the system of inequalities:
1) 2) 3) 4)
Create an inequality corresponding to the condition of the tables and solve it.
1) x at what values of y= 0,5x+2 and y= 3-3x the values of the functions are simultaneously: 1) positive; 2) negative; 3) 3 and older; 4) is less than 3?
2) x at what values of y=x-2 is going y= 0,5xThe values of the functions +1 are simultaneous: 1) negative; 2) nomusbat; 3) not less than 4; 4) is not greater than 4?
3) One side of the triangle is 5 m and the other side is 8 m. The perimeter of a triangle is: 1) less than 22 m; 2) If it is more than 17 m, what can be the third side of it?
4) If a part of an integer is part of it, then a number greater than 29 is formed, if part of the same number is subtracted, then a number less than 29 is formed. Find this whole number.
5) If half of the whole number is added to the double, then a number less than 92 is formed, if half of the same whole number is doubled, then a number greater than 53 is formed. Find this whole number.
MMIBDO ': / / B.Teshaboyev
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Topic 17: The Greatest Common Divisor (EKUB)
If, EKUB (a, b) = 1, the numbers a vab are called mutually prime numbers.
For example: (1; 2), (2; 3), (15; 28), (10; 21) and so on
- If a = 2² ∙ 5² ∙ 7 and b = 2 ∙ 5³ ∙ 11, find EKUB (a, b).
Solution: EKUB (a, b) = 2 ∙ 5² = 50
- Find the EKUB (345, 285, 315).
Solution: Divide the numbers 345, 285, 315 into prime factors. 345 = 3 ∙ 5 ∙ 23; 285 = 3 ∙ 5 ∙ 19; 315 = 3² ∙ 5 ∙ 7 → EKUB (345,285,315) = 3 ∙ 5 = 15
Let's write all divisors of numbers 24 and 90:
The common divisors of the numbers 24 and 90 are: 1, 2, 3, 6. The largest of these common divisors is: 6.
The number 6 is called the greatest common divisor of the numbers 24 and 90.
The greatest common divisor of the natural numbers m and n is defined as: EKUB (m, n).
So,.
1-for example. Find EKUB (84, 96).
Solution. .
2-for example. Find EKUB (15, 46).
Solution.
The numbers 15 and 46 have no common prime divisors. In such cases, the largest common divisor of the given numbers is 1. So for the numbers 15 and 46.
1. The winners of the competition in mathematics will be awarded with notebooks and pens. How many notebooks and pens will be given to each winner out of 42 notebooks and 30 pens? What is the maximum number of winners?
Solution: We find the common divisors of 42 and 30.
They are: 1,2,3,6, so the number of winners can be the same. The largest is 6 J: 6.
- EKUB (720, 540) =?
Solution: 720 = 2 ∙ 3² ∙ 5 and 540 = 2² ∙ 3³ ∙ 5
EKUB (720,540) = 2² ∙ 3² ∙ 5 = 180 Answer: 180
MMIBDO ': / / B.Teshaboyev
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Topic 18: SMALL GENERAL QUARTERLY
The winners of the mathematics competition will be awarded with notebooks and pens. How many notebooks and pens will be given to each winner out of 42 notebooks and 30 pens? What is the maximum number of winners?
Solution: We find the common divisors of 42 and 30.
They are: 1,2,3,6, so the number of winners can be the same. The largest is 6 J: 6.
- EKUB (720, 540) =?
Solution: 720 = 2 ∙ 3² ∙ 5 and 540 = 2² ∙ 3³ ∙ 5
EKUB (720,540) = 2² ∙ 3² ∙ 5 = 180 Answer: 180
Let's write multiples of 36 and 48:
Among these numbers are numbers that are common to both rows:
144, 288, 432, ...
They are the common multiple of the numbers 36 and 48.
The common multiple of numbers divisible by 36 and 48 is: where k is an arbitrary natural number.
But the number 144 is the smallest of all numbers multiplied by 36 and 48. We call the number 144 the smallest common multiple (divisor) of the numbers 36 and 48.
Hence, EKUK (36, 48) = 144.
Here are two ways to find EKUK.
1-for example. Let EKUK (15, 12) be found.
Method 1 The largest of the numbers is 15. Let's write multiples of it and find out whether they are divisible by 12 or not:
the number is not divisible by 12, the number is not divisible by 12, the number is not divisible by 12, the number is divisible by 12.
Hence, EKUK (15, 12) = 60.
Method 2 Divide the numbers 15 and 12 into prime factors:
and.
The number EKUK (15, 12) is a number divisible by both 15 and 12. Therefore, all non-common multipliers of the numbers 15 and 12 are also involved in its propagation. The common prime multipliers are taken from one.
So,.
Example 2 Let EKUK (20, 33) be found.
and -relative prime numbers, they have no common prime divisors.
In that case, it will
- Find the ECU of 48 and 60.
Solution: 48 = 2 ∙ 24 = 2 ∙ 3 60 = 15 ∙ 4 = 2² ∙ 3 ∙ 5
EKUK(48,60)=2∙ 3 ∙5=16 ∙15=240
- Find the ECU of the numbers 24,35, 74 and XNUMX
Solution: 24 = 3 ∙ 8 = 2³ ∙ 3 35 = 5 ∙ 7 74 = 37 ∙ 2
EKUK (24, 35, 74) = 2³ ∙ 3 ∙ 5 ∙ 7 ∙ 37 = 31080
a) They want to sell the fabric from 4 meters or 5 meters. How many meters of fabric should be at least to prevent clotting?
Solution: We need to look for a number that is divisible by 4 and 5.
This is the EKUKI of 4 and 5. EKUK (4, 5) = 20
Answer: 20 meters
b) The product of two numbers is 294, and the greatest common divisor of ul is 7. Find EKUK for these numbers.
Solution: Since EKUB (a, b) EKUK (a, b) = ab EKUK = 294: 7 = 42
MMIBDO ': / / B.Teshaboyev
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Topic 19: Two numbers are their sum and division (ratio) find on.
The basic issue. The sum of the two numbers is 200. one number is 3 times larger than the other, find these numbers.
Solution: small number 1 part Large number 3 parts Total 4 parts
To find a small number, we divide 200 by 4; multiply the obtained division by 3; we find a large number.
We add both numbers to check
- 200: 4 = 50;
- 50 3 = 150;
Control: 50 + 150 = 200
Issue 1. If the rest of the day is five times more than in the past, what time is it now?
Solution: the sum is 24 and the division is 5. This means that the previous part of the day is equal to the hour and the rest is equal to the hour.
Issue 2. The age of the mother is three times the age of the daughter, and the age of the father is the same as the age of the mother and daughter, if the sum of the ages of all three is equal to the sum of four If so, how old are each of them?
Solution: age of all together: 100 + 4 = 104 years. The mother has three parts, the daughter one part and the father 3 + 1 = 4 parts. All of these pieces were 3 + 1 + 4 = 8.
So, the age of the daughter: da; mother age; father
year old. The following problems can be solved in the same way.
Issue 3. The sum of the two numbers is 410, and when a large number is a small number, it is multiplied by 7 and multiplied by 10. Find these numbers.
Issue 4. The division of two numbers is 3 and the remainder is 10. If the divisor, the divisor, the divisor and the remainder are added, it is 143. Find the divisor and the divisor.
MMIBDO ': / / B.Teshaboyev
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Topic 20: The difference and division of two numbers (ratio) to find on.
The basic element is that the father is three times older than the son. If a father gives birth to a son when he is 24 years old, how old is each of them?
If we subtract 1/3 of the age of the son from the number of the age of the father, then 2/3 of the age of the father remains, which is 24 years. We find the whole number by the fraction of the number: age.
Issue 1. My brother and sister had money. If a brother gives 24 rubles to his sister, their money will be equal to each other, if a sister gives 27 rubles to her brother, her brother's money will be twice as much as her sister's. How much money did each of them have?
Solution: 1) His brother has 48 soums more than his sister
2) If a sister gave 27 soums to her brother, the difference would increase to 54 soums and would be 102 soums (48 + 54);
3) At that time, his brother had twice as much money as his sister. We determine how much money a sister should have in terms of difference and ratio: soums;
4) After giving 27 soums to her brother, her sister has 102 soums left. So, before he had 129 soums (102 + 27);
5) His brother had more than 48 soums. So, his brother had 129 + 48 = 177 soums.
Issue 2. One boy said to another, "Give me an apple, and I will have twice as much as you." The other replied, "No, you give me one apple, and we will have two." How many apples did each of them have?
Solution: 1) It is clear from the words of the second child that his apple is two times smaller than that of the first child.
2) If the second child gave another apple to the first, the difference would be two more and would be equal to 4.
If the second child gave an apple to the first child, he would have an apple. So his apple is 4 + 1 = 5. The first is 5 + 2 = 7.
The following issues can also be solved in this way.
Issue 3. There are two freight cars at the railway station. (all wagons are the same length) The number of wagons in one convoy is more than 12 in the second convoy; After separating 4 wagons from each of the two trains, the first train was twice as long as the second train. How many cars were in each train?
Issue 4. When the boy was asked how many brothers and how many sisters he had, he replied, "The more brothers I have, the more sisters I have." When his sister was then asked how many brothers and sisters she had, she replied, "My sisters are twice less than my brothers." Is that possible?
MMIBDO ': / / B.Teshaboyev
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Topic 21: Finding two numbers using their sum and difference
Basic problem: if the sum of two numbers is 1000 and the difference of these numbers is 292, find these numbers.
Since a large number is a small number + a difference, the sum of the two numbers can be thought of as the addition of the difference to the doubling of the small number.
After subtracting their sum from the sum of two numbers, we get a double of a small number. If we add the difference to the sum, we get a double of a large number.
Method 1: 1) 1000 - 292 = 708
2) 708: 2 = 354 (small number)
3) 354 + 292 = 646 (large number)
Check: 354 + 646 = 1000.
Method 2: 1) 1000 + 292 = 1292 2) 1292: 2 = 646 (large number)
3) 646 - 292 = 354 (small number) Check: 354 + 646 = 1000.
Issue 1. Three bags of potatoes weigh 156 kg. The first bag is 18 kg heavier than the second, and the second is 15 kg lighter than the third. How many potatoes are in each bag?
1) (kg) 2) (kg)
3) (kg) Check: 59 + 41 + 56 = 156 (kg)
Issue 2. The mother was 32 years old when her daughter was born and 35 years old when her son was born. If the age of all three is 59, how old is each now?
Solution: The youngest is his son. Her sister is older than her (35-32). The mother is 35 years older than her son. The son is old. Her daughter is 10 years old. His mother is 42 years old.
MMIBDO ': / / B.Teshaboyev
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Topic 22:Solve speed detection problems.
Elementary matter. The ship traveled along the water at a speed of 20 km per hour, against the current at a speed of 15 km per hour. Find the velocity of the water.
Solution: the speed of the ship along the current is equal to the sum of the speed of the ship and the speed of the current; and the velocity as it travels against the current is equal to the difference. It turns out that the difference between the speed of the ship in the current and the speed against the current is equal to the double speed of the current.
This means that the speed of water is km.
Issue 1. The boat can swim at a speed of 7 km per hour. It takes less time to swim the distance between two points than it does to swim against the current. Find the velocity of the water flow.
Solution:
The boat can travel the distance MC in 1 hour, of which DC = 7 km is the speed of the boat, and MD is the speed of the boat. Similarly, the distance AB, the boat travels counterclockwise. If there was no current, it would have traveled a distance AN = km greater than that per hour. The boat can cover this distance in 1 hour due to its movement (CD = BD = 7 km) and per hour due to the flow of water (MD = AD and BN = AD). This means that the speed of water flow per hour is: km, ie the speed per hour is km.
The following issues can be included in the same type.
Issue 2. Water flows at a speed of 3 km per hour; It takes 3 times less time for a boat to travel a certain distance along a stream of water than to swim against the current. Find the speed of the boat in still water.
Issue 3. As the boat traveled along the water, it passed between two points in an hour. On the way back, he covered this distance in 6 hours. How long did it take for the thrown timber to travel this distance along the stream?
MMIBDO ': / / B.Teshaboyev
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Topic 23: Meeting Movement Issues.
Elementary matter. The distance from the village to the city is 45 km. At the same time, facing each other, a pedestrian and a cyclist set off. The speed of a pedestrian is 5 km per hour, and that of a cyclist is 10 km. how long will they meet?
Solution:
The distance between a pedestrian and a cyclist is reduced by 10 + 5 (km) per hour. The sum of the speed of a pedestrian and a cyclist is the number of times they meet in 45 km: hours. Answer: They will meet in 3 hours.
Issue 1. One train is passing a train coming from the other direction; the first is moving at a speed of 50 km per hour and the second is moving at a speed of 58 km. A passenger on the first train watched the second train pass in 10 seconds. Find the length of the second train.
Solution. The second train passed the observer on the first train for 10 seconds at a speed equal to the sum of the speeds of both trains. So the length of the second train
Answer: The length of the second train is 300 m.
Issue 2. The distance from Kokand to Margilan is 75 km. At 9 o'clock in the morning the cyclist left Kokand. At 9:36 in the morning, the second cyclist set off from Margilan and traveled less than a kilometer per hour. The cyclists met in the afternoon, how far did they meet from Margilan, how many km did each of them travel, when did the first one reach Margilan?
Solution: The second cyclist traveled less than a kilometer per hour than the first. By the time of the meeting, he was walking for hours. If the first cyclist was walking at the same speed as the second, he would be walking less than 3 hours. This means that if both cyclists were walking at the same speed as the second cyclist, they would have crossed the road. It follows that the speed of the second cyclist is km / h. The meeting was 30 km from Margilan. The speed of the first cyclist was km / h, and he arrived in Margilan 2 hours after the meeting (30:15 = 2), ie at 2 o'clock in the afternoon.
MMIBDO ': / / B.Teshaboyev
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Topic 24: Chasing actions.
Elementary matter. The father sent his son to bring books from the city. But he forgot to say which books to bring. 3 hours later he was chased by a bicycle. If the son travels 5 km per hour and the father travels 8 km per hour, how many hours will the father catch up with his son?
Solution: the son walks 15 km () in three hours, and the father walks more than three km (8-5 = 3) every hour. It will take his father 15 hours (5: 15 = 3) to travel the extra 5 km, ie an hour.
Issue 1. The dog is chasing the fox, but the distance between them is the same as the distance the dog jumps a hundred times. When a dog jumps three times, a fox jumps 5 times, but in terms of length, a dog jumping six times was equivalent to a fox jumping 11 times. How many jumps can a dog chase?
Note: The difficulty of this problem is that both time and distance are expressed in the same unit, that is, by jumping. It is useless to replace these concepts. This difficulty is further complicated by the need to turn the dog's jump into the fox's jump and vice versa.
Now let's see how the problem is solved.
Solution: 1) When a fox jumps 5 times, a dog jumps three times.
So when a dog jumps six times, a fox jumps 10 times.
- A dog jumping 6 times is equivalent to a fox jumping 11 times in length. that is, when the dog jumps 6 times, it approaches the fox in the amount of one jump (in terms of length).
- A fox's 11 jumps are equal to a dog's 6 longitudinal jumps. So one jump of a fox is equal to a dog's long jump.
- The dog approaches the fox as part of its jump in 6 jumps, and as close as part of its jump
- To find out how much a dog can reach a fox when it jumps, divide the dog's 100 jumps by the dog's jump, so that the answer to the question in question is when jumping.
There may be different options for this solution. Here are some of them. Option 1. In its 6 jumps, the dog approaches the fox in the amount of one jump, that is, the dog approaches the fox in one jump. Let's convert a dog's 100 jumps into a fox jump: and that's 1100 dog jumps.
Option 2. The speeds of both the dog and the fox are inversely proportional to the jumps that occur at the same time (), meaning that the speed of the dog is the same as the speed of the fox. This means that the dog gets closer to the fox every time he jumps, chasing the fox in the jump.
Option 3. 1) The dog approaches the fox in six jumps, the size of which is one jump.
2) The dog crosses the path () up to 66 jumps of the fox in 11 jumps.
3) Two jumps of a fox are equal to 6 jumps of a dog. This means that the dog travels in its 66 jumps more than the fox in its 6 jumps.
4) the dog approaches the fox in one jump in the amount of two jumps.
5) The dog chases the fox in 1100 jumps.
Issue 2. The pedestrian walked from A to B. After 12 hours, the car left from A to B. A car travels 5 times faster than a pedestrian. In how many hours will the car catch up with the pedestrian?
Solution: a pedestrian walks the road in 12 hours, a car in 5 times less time, ie in an hour. Assuming that the speed of the car is 1 and the speed of the pedestrian, the car is approaching the pedestrian at its own speed every hour. The distance traveled by a pedestrian in 12 hours is expressed by the speed of the car, which is equal to, and the car catches up with the pedestrian 3 hours after he started walking, or 15 hours after walking (12 + 3 = 15).
MMIBDO ': / / B.Teshaboyev
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Topic 25: Replacing one quantity with another.
Elementary matter. 8 m of satin and 5 m of chit cost 835 soums. If one meter of satin is 1 soums more expensive than 28 meter of chit, how much does each meter of satin and chit cost?
Solution: 1) If we bought 8 m of satin instead of 8 m of satin, we would have saved 28 soums for each meter of satin, and the total would have been reduced by 835-224 soums. = 611 soums.
2) 13 meters (8 + 5 = 13) fence would cost 611 soums, and 1 meter fence would cost 611: 13 = 47 soums.
3) One meter of satin costs 28 soums per meter, that is, one meter of satin costs 47 + 28 = 75 soums.
Let's look at solving more complex problems.
Issue 1. cubic meter of dried apricot wood and cubic meter of dry spruce is t, and one cubic meter of dry apricot is heavier than one cubic meter of spruce. What is the weight of a pine tree and a cubic meter of spruce?
Solution: 1) replace the spruce with apricot wood. If an apricot is several times heavier than a spruce, the size of an apricot tree, the weight of which is equal to the weight of a spruce tree, is part of the size of a spruce, ie cubic meters. ladi.
2) cubic meters and cubic meters of apricot wood t, and one cubic meter of apricot wood t and one cubic meter of spruce t.
Issue 3. 32 m of chit, 40 m of satin, 25 soums were sold for 4998 soums. If one meter of yarn is 2.4 times more expensive than one meter of chit, and one meter of satin is 1.44 times cheaper than one meter of yarn, how much does each meter of chit, satin and yarn cost?
Solution: 1) Replace the screw with a chit. The swing is 2.4 times more expensive than the fence, which means that instead of 25 meters of swing you can get 2.4 times more swatches (you can get 25 times more switches for the money paid for 2.4 meters of swing), ie.
2) replace the satin first on the stitch, then on the edge. Satin is 1.44 times cheaper than surup. So, for the money paid for 40 m of satin it is possible to buy 1.44 times less, that is 40 m: 1.44 = m. It is possible to buy a fence for 2.4 times more than the money paid for the drive.
3) For 4998 soums you can buy a fence in total, so one meter of fence costs one meter, one meter of satin costs 75 s. 60 t: 1.44 = 52 s. It costs 50 t.
MMIBDO ': / / B.Teshaboyev
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Topic 26: ye number history
This number appeared recently. It is sometimes called the "neper number" and is associated with the name of the Scottish mathematician John Nepera (1550-1617), which is unfounded because Neper ye not sure if you have a clear idea of the number. «ye»Designation was introduced by Leonard Euler (1707-1783). ye found 23 numbers using the infinite series expression of. »In 1873, Hermit proved that ye were a transcendental number. L.Eyler eat and found a wonderful relationship between. ye The logarithms on the basis are considered and Lx is defined as
eat sdecimal digits of the moment
e = 2.718281 8284590452 3536028747 1352662497 7572470936 9995957496 6967627724 0766303535 4759457138 2178525166 4274274663 9193200305 9921817413 5966290435 7290033429 5260595630 7381323286 2794349076 3233829880 7531952510 1901157383 4187930702 1540891499 3488416750 9244761460 6680822648
MMIBDO ': / / B.Teshaboyev
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Topic 27: Equalize the data and subtract one from this.
Elementary matter. 400 soums were paid for a kilo of biscuits with 144 g of candy. In another purchase, the same 600 g of candy was paid 136 soums per kg of oven. How much is a kilo of candy and a kilo of biscuits?
Solution: 1) Let's equate one of the two given quantities: 1200 g of candy costs 432 soums per kg of biscuits, 1200 kg of biscuits with 2 g of candy costs 272 soums.
2) This means that the difference between the price of candy and biscuits (432-272 = 160 soums) depends only on the difference between the amount of biscuits purchased.
3) find the price of the biscuit. som
4) One kg of biscuits costs 64 soums, 600 g (in the second purchase) costs 136-64 = 72 soums and one kg of candy costs one soum.
Issue 1. 4365 kg of rice was delivered to two shops: one part of rice delivered to one store and one kg of rice delivered to another store. How much rice is delivered to each store?
Solution: Store I is listed in Store II
All from the first row
Let's separate the second rice
Half of the last row
The last two rows
sum
The last one from the second row
we separate
So, the rice brought to the second store:
506 kg: = 1518 kg
The rice brought to the first store is:
4365 kg - 1518 kg = 2847 kg
Issue 2. The bow has three banknotes and 5 banknotes of 50 soums. If it was twice less than three soums and three times less than 5 soums, the number of both types of money would be 19. How much money do you have in your pocket?
Solution: In order to prevent fractions, we solve the problem according to the second condition (there are 19 banknotes in the pocket; if we double the number of three rubles and triple the number of 5 rubles, the number of banknotes will be 52 ladi), then the problem is solved as follows:
Number of 3 soums + number of 5 soums = 19;
Double number of 3 soums + Double number of 5 soums = 38;
Double number of 3 soums + triple number of 5 soums = 50.
Equating the last two equations, we find that in the first case 5 sums are 50-38 = 12, but this is 1/3 of what is in the pocket, so 5 sums is 123 = 36; 3 soums were 50-36 = 14.
MMIBDO ': / / B.Teshaboyev
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Topic 28: Collaborative work.
Elementary matter. One worker completes a task in an hour, and another in 5 hours. How many hours will both workers complete the job?
Solution: 1) The first worker did all the work in an hour, and less than once an hour, that is, part of the work.
2) the second worker does part of the work in an hour.
3) When the two work together, they do part of the work in an hour.
4) and all finish the work in 3 hours (1: 1/3 = 3).
Issue 1. The pump delivers 900 liters of water to the pool per hour. When the pump is running continuously, all the water flows through the first pipe in 12 hours, and through the second in 10.5 hours. When both the pump and the pipe were turned on, the pool was drained in 5 hours. Find the size of the pool.
Solution: 1) Part of the full pool and 900 liters of pumped water flow per hour through the first pipe, and part of the full pool and 900 liters of pumped water flow from the second pipe.
2) 900 liters2 = 1800 liters of water supplied by a part of the full pool and the pump per hour through both pipes; At 5 hours, part of the full pool and 1800 = 9000 liters of water supplied by the pump will flow.
3) 5 liters of water comes through the pump in 4500 hours. This means that in 5 hours, a full pool of water and 4500 liters of water flow through both pipes; This will be part of the pool and will be 9000 liters, that is, part of the pool will be 4500 liters.
4) Now we find the whole number by the fraction of the number: the volume of the pool is 4500 liters: = 42000 liters.
MMIBDO ': / / B.Teshaboyev
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Topic 29: Finding two multipliers with the help of their given multipliers and differences when their products are equal.
Elementary matter. A few chickens and a few geese were bought for the same amount of money, but 20 more chickens were bought than geese. One goose costs 126 soums and one chicken costs 70 soums. How many geese and how many chickens were bought?
Solution: 1) More than 20 chickens cost 1400 soums (70 20 = 1400 soums). How did this money come about? When buying one chicken and one goose, 56 soums (126 soums - 70 soums = 56 soums) were spent less per hen than one goose. The same savings were made when buying a second chicken and goose. So, before 1400 harvest, they kept the same economy and bought 1400 extra chickens for 20 soums.
2) So, the more geese are bought, the more times there are 1400 soums in 56 soums, 1400: 56 = 25. Thus, 25 geese are bought, and chickens are more than 20, ie 45 chickens were obtained.
A complicated matter. The train covered the distance between the two stations in 2 days, traveling for 3 hours every day. If a train travels every day for 18 hours and 22 minutes and travels more than 30 km per hour, how many days will it take to cover this distance?
Solution. 1) The distance between stations is 54 hours (183 = 54) with normal train travel. If the train increased its speed by 11 km per hour, it would cover this distance in 45 hours, ie 9 hours ago.
2) If the train was traveling at a higher speed, it would cover an additional 45 km in 495 hours, which would take an additional 9 hours to cover this distance in normal travel.
3) This means that the normal speed of the train is 495: 9 = 55 km / h, and the distance between the stations is 55 km 54 = 2970 km.
MMIBDO ': / / B.Teshaboyev
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Topic 30: Problems to be solved from the end.
Elementary matter. There were a few apples in the box. The first child received a quarter of the apples in the box and 3 more. The second child took one-third and 4 of the remaining apples. The third child received half of the rest and 6 more. Then there are 2 apples left in the box. How many apples were in the box and how many apples did each child get?
Solution. This type of problem will be easier to solve from the end.
1) There are 2 apples left in the box, before that the third child got 6 apples and half of all the apples left in the box before that. It turns out that the third child took half of the apple in the box. The second half, equal to 8 apples (2 + 6 = 8), remained in the box. So the third child got 8 + 6 = 14 apples and there were two apples left in the box. Thus, the second child had 16 apples left in the box.
2) The second child received 4 apples, followed by 16 apples. So, after the second child gets all the remaining apples, there is a piece of apple or 20 apples left in the box. He bought all the apples, ie 10 apples and 4 more apples - a total of 14 apples; then 16 apples remained. So, after the first child, there are 30 apples left (14 + 16 = 30).
3) The first child received three apples and a portion of all the apples in the box before. When he took part, there were 33 apples (3 + 30 = 33) left in the box. He took a portion of all the apples, 11 apples (33: 3 = 11) and 3 more apples, for a total of 14 apples, and in the box were 44 apples (114 = 44). .
MMIBDO ': / / B.Teshaboyev
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Topic 31: .Interesting and various life situation issues
Issue 1. There are bacteria in the glass. After one second, each of the bacteria splits into two equal parts, then each formed bacterium splits into two equal parts after one second, and so on. After how long will the glass be half full?
Answer. After 59 seconds.
Issue 2. Anya, Vanya and Sanya got on the bus, which had no small copper coins, but paid the fare. they pay five cents each. How did they do it?
Solution. Anya and Vanya paid Sanya 15 shillings, of which 10 shillings were returned. After that, he paid 15 cents.
Issue 3. Part of the book has fallen off, its first page It has a serial number of 328, the last number being written with the same numbers but in some other order. How many pages are there in the dropped section?
Answer: 495 pages
Issue 4. The bag contains 24 kg of nails. How to pull a 9kg nail without having a scale without miles?
Solution. First we divide the nails into two equal parts - from 12 kg to two groups, then we divide one of these groups into two equal parts, then again we divide into two equal parts. We remove the obtained 3 kg of nails and take the remaining 9 kg. .
Issue 5 The slug crawls along the column from its base, every day it falls 5 cm up and 4 cm down every evening. If the height of the column is 75 cm, when will it reach the end of the column?
Solution. The muskrat will be at the end of the column on the evening of the 71st day.
Issue 6 In January of a year there were four Fridays and four Thursdays. What day of the week was the 20th of this month?
Answer: Sunday.
Issue 7 How many rooms intersect diagonally in a rectangle with dimensions 199 × 991?
Solution. Diagonal 199 + 991 - 1 = 1189 intersects the room.
Issue 8. Delete 1234512345123451234512345 numbers from the number 10 so that the remaining number is the maximum possible number.
Answer: Maximum number is 553451234512345.
Issue 9 Petya said, “Before yesterday I was 10 years old, next year I will be 13 years old.” Could that be so?
Solution: yes, it is possible, if Petya's birthday is December 31, and he said this on January 1.
Issue 10 Petya's cat sneezes all the time before the rain. Today he sighed. "So it's going to rain," thought Petya. Is he right?
Answer: No, it's not true.
MMIBDO ': / / B.Teshaboyev
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Topic 32: Numerical History
The history of numbers dates back to the Egyptian papyri of 2000 BC. but it was also known to the ancients. Since then, natural numbers 1,2,3,4,… have been inseparable companions of human thought, helping to determine the number of objects or their lengths, surfaces or volumes. acquainted with the number of people as given. At that time it was not marked with any letter hyech of the Greek alphabet and its role was performed by the number 3. It is not difficult to understand why so much attention has been paid to numbers. Expressing the amount of the relationship between the length of a circle and its diameter, it appeared in all matters relating to the face of a circle or the length of a circle. ' But even in antiquity, mathematicians found that the number 3 was not as accurate as the number pi. Obviously, they came to this only after the appearance of fractions or rational numbers among the natural moments.
Archimedes found other limits of the number pi using the method of upper and lower approximations. The designation of the number began to be used systematically after Leonard Euler began to use it systematically in the late eighteenth century. Legendre proved to be an irrational number. In 1706, F. Liderman proved that he was a transcendent, that is, he did not satisfy any algebraic equation with any coefficient.
During the entire existence of the number, a peculiar chase was carried out to find the numbers of its decimal rooms. Leonard Fibonacci in 1220 determined his three correct decimal numbers. In the 16th century, Andrian Antonis found 6 such numbers. François Viet (like Archimedes, he found 322216 exact numbers by calculating the perimeters of the interior and exterior angles 9. Van Kyolen calculated the perimeters of angles 15 and calculated 1073741824 exact numbers. Abraham Sharp found 32512254720 exact numbers. In 20 Z. Daze found 72 post-comma numbers. In 1844 T. Clausen found 200 numbers and in 1847 Richter 248. Z.Daze found the number 1853 in 330, and U.Shenks found the number 1853 in the same year.
1949 - 2037 decimal places (John von Neumann, ENIAC),
1958 - 10000 decimal places (F.Jenyui, IBM-704),
1961 - 100000 decimal places (D.Shenks, IBM-7090),
1973 - 10000000 decimal places (J. Giyu, M. Buye, CDC-7600),
1986 - 29360000 decimal places (D. Bailey, Cray-2), sdecimal digits of the moment
= 3.1415926535
MMIBDO ': / / B.Teshaboyev
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Topic 33: Problems to be solved by assumptions.
Elementary matter. The farm has chickens and sheep. If they all have 19 heads and 46 legs, determine the number of chickens and sheep.
Solution . 1) Suppose there are only chickens on the farm. They would have 38 legs (219 = 38). In fact, the number of legs is not 38, but 46, ie more than 8. Why. Because when we replace sheep with chickens, we reduce the number of legs in each sheep by 2 (4-2 = 2), so we have 8 less legs. This means that if there are more than 8 out of 2, the number of sheep on the farm is the same.
sheep.
2) We can assume that there are only sheep on the farm. Then they would have 79 feet (419 = 76) and nfyotsh would be 30 feet more than they really are. When we exchange chickens for sheep, we add two legs to each hen, for a total of 30 legs. If there are more than 30 out of 2, the number of chickens on the farm will be the same. 30: 2 = 15 chickens.
Issue 1. The shopkeeper sold 95 kg of 3 types of sugar: 1 kg of 1 type 137 s. 50 tiyn, the second type - 2 soums, and the third type - 135 soums. If the total amount of sugar sold is 3 soums, and the 124st type is sold 12730 times more than the 1nd type, how many kg of each type will be sold?
Solution: 1) 2 kg of the 1nd type corresponds to 2 kg of the first type. So, with 2 kg, one kg of type 2 costs 137.5 s 2 + 135 s = 410 s, and one kg of a mixture of these two types costs 410: 3 = soums.
2) Suppose that all 95 kg of sugar is of the 3rd type, in which case sugar was 124 s 95 = 11780 soums, ie 950 soums less than the amount paid for all sugar (12730-11730 = 950). This is due to the fact that we have reduced the price of one kilogram of sugar from the first type to the second.
3) How many times in 95, the first type is sold as much as the second type of sugar: kg.
4) The first type of sugar is sold twice as much as the second type. kg kg of the first variety, kg of the 3rd variety were sold.
Issue 2. 2380 tons of cement were purchased for 435 soums per ton. Part of this cement is brought in with the tin and part in the barrel. There is t cement in both a bag and a barrel. 1263900 soums were paid for cement, bags, barrels, 100 soums for each barrel, 75 soums for each bag, how much cement was delivered in bags, how much in barrels ?
Solution: 1) 2380435 = 1035300 soums were paid for pure cement.
2) money paid in barrels with a bag
1263900-1035300 = 228600 y
3) The total volume of the bag and barrel was 6435 = 2610.
4) If all the dishes consisted of barrels, it would cost 1002610 = 261000 soums.
5) in fact it is cheaper by 32400 soums
(26100-228600 = 32400)
because a bag is not 100 soums, but 75 soums, ie 25 soums cheaper.
6) If there are 32400 times in 25 soums, the number of bags will be the same 32400: 25 = 1296, including 1296: 6 = 216 tons of cement.
7) barrels are 2610-1296 = 1314 and the cement in them is 1314: 6 = 219 t.
Answer: 216 tons of cement in cans and 219 tons in barrels.
MMIBDO ': / / B.Teshaboyev
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TOPIC 34: Math Night Event
Beginner
Peace be upon you, O blessed ones of the earth
Free generations of good-natured people
We saw each other on a good day
Is there more happiness, my dears.
Dear moments of our dear age
Dear people ask dear
Opportunity is a trophy with royal lines
It is time to decorate the notebook of life
In fact, no matter what era we are in, we start our first word with an Uzbek greeting. Because this is one of the mysterious and simple aspects of etiquette for us.
First of all, we would like to welcome all the members of the team participating in this competition, our spectators and our coaches, who share the beauty of our circle and give it beauty.
The main purpose of this competition is: 5 "A" - students of the class compete in math with their friends. It is to further enhance the knowledge and skills we have acquired to date.
This is our five-day life,
It flows like water.
The day we saw yesterday,
Today is left behind.
Sometimes we cry, sometimes we are happy
Sometimes we repent, sometimes we are free
Every day is different
Life is so lifelong.
Al-Khwarizmi and Al-Beruni will compete in today's quiz "Let's test our knowledge". Each group consists of 15 students, the terms of which are as follows:
1 - condition. Introduction
2 - condition. Question and answer
3 - condition. Mutual question and answer of groups
4 - condition. Competition of group leaders
Please watch.
We offer the first group, Al-Khwarizmi, on the first condition.
Local:
Peace be upon those gathered here
Dear friends, dear ones
Our people are raising a child
To wise and knowledgeable teachers
Assalamu alaykum, dear masters who dedicate their hearts to the younger generation are dear peers of our country.
On behalf of the students of our group, we wish good luck to you dear teachers and peers.
Our homeland: Uzbekistan
Our city: Beautiful Navoi
Our motto: Exemplary behavior and excellent reading
The foundation you put on the science of algebra
Khorezm is a teacher for you
The universe is your mind full of meaning
It is clear that the face of accounting science is true
How long will your captives live?
It will be priceless as time goes on
Our goal is to study it deeply
Muhammad Musa Al-Khwarizmi is one of the great scholars of his time. Al-Khwarizmi was born and raised in the land of Khorezm in 483. He wrote many works. Ten of his works have arrived.
Centuries to come, centuries to come
But the basics of science created
Proverbs from century to century
Our ancestor was the meat of the Khorezmian spirit
I am glad to hear your words
We know there are 10 numbers in math, so let’s listen to their conversation.
Fahriddin:
I have a math job
Suddenly the previous zero is forbidden.
If I come later
You can add a dozen
Dilshod:
Mendirman is an end
Life begins with me
Although I am small and odd
Each issue has a ven hamdam
Shahijahan
I add two bans
The captain of even numbers
I am three or four years younger
But they are tired
Malika
Number three
Assessing my knowledge
Noiloj is satisfied
Sometimes I find myself five
I'm tired of staring
Prince
The number four
If you know a friend
Don't upset me
If you envy that four events
Let's just add one to three
Basil;
Number five
They call me the number five
The soul of the elite
Three is less than me
It's six o'clock, brother
Mohidil:
Number six
Coptoximon belly
I'll take an umbrella
One - two - three to myself
I can be divided equally
Gulshada;
The number seven
Wearing a hat on my head
I fastened the belt
I am ready to serve
Mehmatsevar comrade
Prince
The eighth issue
Silent - I have a sleek shape
What you see is what you want
Clean and beautiful too
If you learn to write
Muhammadjan:
The ninth issue
I'm nine, you know
Learn to count quickly
You add two to seven
One less than eight
Yunusbek:
Add
I'll add it one by one,
I add strength to the numbers.
My belt line
I stand on my own
Lily: Reproduction
Multiply the numbers
Increase several times
I admire my work
Reproduction is good
Being a lover
If the numbers increase
I'll give it to you
An example if you work
I will be two points
Al-Beruni
Assalamu alaykum dear teachers and dear spectators. Thank you so much for this competition and for your visit. We start our introductory part by wishing you good luck during the competition
Our name: Algebra
Our goal: To uncover the undiscovered aspects of the field of mathematics
Our motto: Good reading.
In life: to be patient
At school: Walking in honor.
In the future: Achieving dreams
However
To the judges: justice
To the audience: patience
rival group: happiness
To ourselves: good luck, and good luck again!
Rahmonali: If you get five marks,
Life will be so beautiful.
: "I don't agree with you, my friend."
"2" will get me all envious.
Shodia: Oh, my friends, I am against this
I get a "1" point, but life is good anyway.
Cameron: Whatever the rating
The name "talented" should not be tarnished!
Condition 2: Questions and answers.
Questions:
- The more sisters a child has, the more brothers he has. His sister has twice as many sisters as his brothers. How many boys and how many girls are in this family.
- The straight line divides the numbers in the clock into two groups. How to draw a straight line so that the sum of the numbers in the two groups is the same.
- Prove that the difference between a 3-digit number and the number formed by writing that number in reverse order is 99.
- There is such a thing as it will be green on the street, black on the market, red on the house. What is this?
After 2 questions for both groups, Al-Khwarizmi left the scene until the judges counted the points. It was attended by: Jurabek, Shahzod.
The question-and-answer session continues after the scene is over.
- Find the largest possible number using three identical numbers.
- Ducks and sheep are walking in the meadow. They all have 30 heads and 84 legs. How many ducks and how many sheep are there in the meadow?
- It's something that can't be eaten, but it can be eaten, it can't be worn, it can be worn. It is as light as a butterfly's wing, but it can subdue things that weigh tons. What is it?
- When Pythagoras was asked, "How many disciples do you have?" He replied. "Half of my students study mathematics, a quarter study nature. Seven of them spend their time meditating, and the rest are three girls." How many students did Pythagoras have?
Condition 3. The groups asked each other questions.
Condition 4. Team Leader Competition.
MMIBDO ': / / B.Teshaboyev
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