TECHNİQUES THAT MAKE STUDENTS THE SUBJECT OF HEURİSTİC ACTİVİTY İN THE PROCESS OF TEACHİNG MATHEMATİCS Текст научной статьи по специальности «Математика»

TECHNÍQUES THAT MAKE STUDENTS THE SUBJECT OF HEURÍSTÍC ACTÍVÍTY ÍN THE

PROCESS OF TEACHÍNG MATHEMATÍCS

Ibragimov F.,

Sheki branch of ASPU, doctor of pedagogical sciences, professor ORCID: https://orcid org/0000-0002-0775-1048

Karimova M. heard teacher at Sheki branch of ASPU DOI: 10.5281/zenodo.6806990

Abstract

The article explains the relevance of the research topic and emphasizes that the level of effectiveness of teaching this subject depends on improving the organization and management of the analytical and heuristic activities of students in the process of teaching mathematics in a secondary school.

The study focuses on the fact that the heuristic type of mental activity is controlled by selecting and systematizing the tasks presented to students. Depending on the use of these means (questions, tasks, exercises) in this process, the functions they perform create various forms of thought movement, such as methods and techniques. These forms of intellectual movement have a positive effect on the formation and development of students' heuristic experience.

Keywords: reception; heuristic technique; receiving small change; proof of the opposite position; removal of excess; auxiliary unknown; assembly and disassembly of the whole into parts; rejection of trial and error; monotony; continuity.

Relevance of the research topic. The level of application of analytical and heuristic activities of students has a significant impact on improving the efficiency of the process of teaching mathematics in a secondary school. An analysis of scientific sources and the experience of practicing teachers indicates that the issue of including the analytical type of mental activity in the pedagogical process, both theoretically and technologically, has been resolved to a greater extent than the heuristic type. Among the reasons for this reality, a special place is occupied by the fact that analytical and heuristic activities are not brought together as sides of a single whole, but are closely related to the creative process. Therefore, when solving the problem of determining the theoretical and technological directions for the application of heuristic activity in the pedagogical process, including in the process of teaching mathematics (this is an eternal problem), it is necessary to study the role of techniques and determine their content. Based on this consideration, we affirm the relevance of the topic of the article.

Interpretation of research material. As you know, the specific activity of the subject for the purpose of learning is learning. This process includes: 1) obtaining information about the essential properties of the world necessary for the successful organization of one or another type of ideal-practical activity; 2) mastering the methods and operations that make up this activity; 3) master the rules for using this information for the correct choice and control of methods and operations that correspond to the conditions of the issue and the upcoming goal. Thus, learning is possible where human activity is directed by conscious goals of acquiring certain knowledge, skills and abilities [5; 192]. Learning is an important characteristic of learning activities and does not cover all of its aspects. includes the acquisition of skills and habits. However, mastery and pedagogical activity are essentially different phenomena. Assimilation is not only a learning

process, but also an integral part of any field of activity [2;150]. The structure of learning activity includes the following components: learning situation (or tasks), learning operation, control, evaluation [7;9]. Learning objectives include students discovering and mastering common solutions to various problems and specific practical issues in their teaching activities. By offering learning tasks, the teacher introduces students to a specific learning situation. There are many features of learning tasks, but one thing is clear: in order to solve them, students must master the necessary learning operations. The structure of learning activity includes the following components: learning situation (or tasks), learning operation, control, evaluation [7; 9]. Learning objectives include students discovering and mastering common solutions to various problems and specific practical issues in their teaching activities. By offering learning tasks, the teacher introduces students to a specific learning situation. There are many features of learning tasks, but one thing is clear: in order to solve them, students must master the necessary learning operations. The structure of learning activity includes the following components: learning situation (or tasks), learning operation, control, evaluation [7; 9]. Learning objectives include students discovering and mastering common solutions to various problems and specific practical issues in their teaching activities. By offering learning tasks, the teacher introduces students to a specific learning situation. There are many features of learning tasks, but one thing is clear: in order to solve them, students must master the necessary learning operations.

Reception is an integral part of the method, or its separate side, in other words, a special concept in relation to the general concept of "method" [1; 275]. Ensuring the cognitive activity of students becomes real only when they master the methods of cognitive activity in a special and systematic way in the process of acquiring knowledge, skills and abilities [8; 42-43]. Mastering these techniques not only has a positive

effect on the formation of knowledge, but also changes the attitude of students to learning, turning it into a purposeful, organized, independent subject [3; 244]. When the structures are focused on the organization and management of analytical and heuristic activities, then their structure (executive steps) is focused on [5; 119]. The methods presented here are aimed at mastering the experience of heuristic activity.

Proof by contradiction. The proof of the opposite proposition is made when it is difficult to prove the opposite directly. The essence of this heuristic approach is as follows. In order to prove the truth A ^ B, the truth B is taken and the attempt is made to prove the truth A.

If it is obtained, then if B ^ A is proved, then the reverse A ^ B is also proved.For example, suppose that 9 a2 - 12 ac +2 b < 0 and we need to prove b < 2 c2. b > 2c2 is accepted in the "opposite position" proof. 9 a2 -12 ac +2 b > 0 proved. In this case, the argument "9 a2 - 12 ac + 2 b < 0 and b < 2c2 - is proven". In other words, b < 2 c2 is considered false, 9 a2 - 12 ac +2 b < 0 turns out to be false.

You can get acquainted with this technique from the 5th grade. To do this, it is necessary to select the appropriate tasks and include them in the educational process. Examples of such assignments include:

1. Prove that if the area of a square is greater than 49 cm2, then its side is greater than 7 cm.

2. Can the area of a square whose side length is expressed as an integer be 201201201201?.

Reception "Changecoin". This method is used to compare values in grades V-VIII and to prove inequalities in higher grades. The essence of the "small change" approach is that a change in a component or model should serve as the basis for obtaining a result, provided that it does not affect the main qualitative characteristics of the object itself. We give examples for clarity.

1-2-3-

... -20

or

1. What's more: 1+2+3+...+1000000?

Solution: 1-2-3-...-20>(2-5) 10-...-20>1012 1 + 2 + 3 + ... + 1000000 <10000001000000 = 1012 So, 1-2-3-...-20> 1+2+3+...+1000000

2. Prove the inequality:

11 11

1001 + 1002 +"' + 2000 >2 Solution: 1 1 1

1001 1002

2000

1 1 1 > + + •" +

2000 2000

2 0 0 0

1000 1 1

=--1000 = -

2000 2

So, ^ + -^+... + -^>1

1001 1002 2000 2 11 11 3. For n > 2--I---+ —I--> - is true. This

n+1 n+2 2n 2

is the general case of the second example problem.

Similarly, if we replace all the sums on the left with the smallest of them, we get: This proves the inequality. —— + — + —+ — > — = -.

^ J n+1 n+2 2n 2n 2

With the "Small change" approach, students can complete the following tasks:

1) What is more ?: a) 635 or 810, b) 156 or 712, 2) Compare: 269 and 1206; 3) Is the inequality true: a) 3111 <1714, b) 1612 < 637

Removing excess. Of the four objects (words, expressions, drawings, etc.) presented in the problems solved in this way, three are essentially similar to each other, and one differs from the others [4; 58]. These tasks require that the excess (which is different from the other three) be identified. To illustrate, consider a few examples.

1. Delete the word excess: plural, difference, stress, part. Answer: The part is not an action component.

2. Remove a longer word: nine, twelve, eight, fifteen.

Answer: 8 is not divisible by 3.

3. Remove the extra expression

a) 310, 27, (-4) 6, (-7) 5.

Answer: (-7) 5 is a negative number.

b) 5ab2, -3 d 2 c; 8 x2 - xy + y; m3

Answer: 8 x2 - xy + y is not one-dimensional.

Separation of the whole part of the fraction.In

teaching "Algebraic Fractions" and "Rational Equations" there is a wide range of opportunities for choosing and applying problems that require this type of heuristic activity, and this is useful [4; 59]. In order for students to actively participate in this type of heuristic search process, attention should be paid to examples that provide specific solutions. This includes:

1.

c 3-8 c+2

find all the complete values of the

fraction that allow you to get the full values c.

Solution. If we separate the whole part from this fraction, we get:

c2 - 2c + 4

16 c + 2

Solving this problem means finding (c + 2), which

gives 16 divisors c.

2. x ^ 0 and y - which of the fractions takes the

10x+y

largest and smallest values for any numbers.-

x+y

Solution. If we apply the division of the integer part of the fraction twice to the given expression, we get:

10x + y 9x 9x 9

X + y

= 1 + -

X + y

= 1 + -

= 1 +

1+ï

y

X + y

X

From this it can be seen that y=0, x when receiving any value, the fraction receives the largest value, i.e. it is equal to 10, and the smallest valuex - = 9 is determined conditionally, in other words, when y = 9, = 1, this fraction takes on the smallest value.x

The following tasks can be used to teach a heuristic approach to extracting the integer part of a fraction.

■In "Algebraic fractions":

1.^+^=7 find the value of the expression

knowing it.—j+T"

2. Find the value of the expression:

a)

b)

*2+3X+7 , r„

-, where x = 53

x+2

7x2-6xy-y2+4

■, where y = 129, x = 6

3. at what full prices n —to determine that the

fraction receives full values-

4. Find the

n—1

smallest

fractions:a) ——,

b)

x2-3x+3 1-x

, if X < 1

5. Schedule a function^ =

-2x2+7x-3

In the teaching "Rational Equations": ■

•v -L 9 „2

■ +

x

= 2

3 (— + — ) — 8

y x

y yx

ß + E wit yjy

+ 10 > 0

If we replace I- + I- with U, then this is

possible ^ + ^ = U2 — 2 and this inequality will take

the form 3U2-8U + 40. The left side of the resulting

1

inequality - < U < 2 is less than zero within the

condition.

JX y

. So this inequality is true.U = I- + I- > 2

The case when the number of variables is saved.

1. Solve the equation:

x2 + x — 5 3x

■+ —-- + 4 = 0

x

x2 + x-5

x2 + x — 5

y =..... " the equation given by the input is a

quadratic equation with respect to the variable y. 2. Solve the equation: (7 — x2)4 + (9 —

x2)4 = 16

y = 8 — x2 after substitution, the equation takes the following form:

or y4 + 6y2 —(y + 1)4 + (y — 1)4 = 16

7 = 0

4x4 + 12x3 — 47x2 + 12x + 4 = 0 let's solve the equation. As you can see, this equation is a return equation (X = 1) and x = 0 is not its root. Divide each side of the equation by x2.

4x2 + 12x — 47 + — + = 2

12 4

4 ( x2 +—~) + 12(x + — ) — 47 = 0

1

x x 1

2

1

x + 1 x2 + 2

■ In the teaching "increasing and decreasing functions":

y = — prove that the function is increasing in the range (-1;^)

Turning on the auxiliary unknown. This heuristic is used to reshape problem text in algebra. The inclusion of auxiliary unknowns can increase, decrease, and remain constant as the initial state changes. Let's give examples of each of them. A situation accompanied by a decrease in the

number of variables.

1. Find natural solutions to the equation: x(x + y) + V*2 + xy + 4 = 52

^x2 + xy + 4 = b > 0 equation given by substitution

can be reduced to b2 + b-56 = 0.

2. Prove the inequality given by the real values of x and y that are different from any zero:

X2) \ X)

if we make a changex + ^ = y 4(y2 — 2) + 12y —47 = 0 4y2 + 12y — 55 = 0 . Considering they are yx = —5,5 ; y2 = 2,5

this when replacing:

a) x + 1 = —5,5, from here 2x2 + 11x +

X

2=0

x1,2 =

-11±V1Ô5

b) x +— = —2,5, from here 2x2

X

1

We get X3 = -; X4 = 2 .

Thus, the roots of this equation

-11+V105 -11-V105 1 , , „ -;-; - and it happens 2.

4 4 2

5x + 2 = 0

3. Solve the system of equations and write down the larger value of y as an answer:

'5(2x — y) 15(x + y) = 28

2x — y

+ 4^3,5*7-1 = 5

X + y 2/3,5*7-2 In the system of equations If we make a substitution ^^ = U, 2/3,Sxy =

x+y

V , we get:

15

5U + — = 28

1 1 9

-V + -F2 = 5 44

o

U = 5, "=§■

V = 4,

V = —5

Since the surface function always gets positive values, it is necessary to subtract V = -5 from the found values of V. Thus, this system is divided into two

systems of equations:

2x-y x+y

= 5,

(2V3,5*y = 4.

(1)

2x-y x+y

5

(2V3,5xy = 4.

(2)

Let's solve the (1) system:

f x + y * a rx + y * 0,

o {2x —y = 5x + ^ o )—3x = 6y,o (3) ( /35xy = 2 (.3,5xy = 4.

f x + y * 0, (x + y * 0, o { x = —2y, o { x = —2^,

l3,5(—2y)y = 4.

2 4

y2 = —.

7

The latter system shows that the set of solutions of (1) is 0.Now let's solve (2).

Îx + y * 0, (x + y * 0,

10x — 5y = 3x + 3y,o { 7x = 8y, o 3,5xy = 4. l3,5xy = 4.

x + y * 0,

o

3,5

_ 8y 7 ,

. 8Z

7

o

y = 4.

(x + y * 0,

8v

x = —, o

7

2

3 „2

x-3

V

3

o

y=-1, . y=1

o

By 7 ,

y=-i, 8Z

7 ,

y=i,

X

X

Thus, this system of equations is a set of

{(-8; -1); 1)} solutions. By condition, we must

indicate the largest of the found answer values, i.e. y = 1

Answer: 1.

Increasing the number of variables.

4^ + 7

n -where the prices are integers ——. Let's assume there is = k , then divisible by 5, because it is. It turns out here too.

4n + 7 = 5fc, 4(n + 3) = 5(fc + 1). There fore n + 3 is divisible by 5 , The biggest common divisor (4; 5) = 1, n + 3 = 5p; peN from here , n = 5p - 3, peW 1. Find the positive root of this equation.

2x + ■

x- 1

1

1---3

1

x--=0

x

If we take the symbols y=J~> z = Vx + 1 then

this equation will take the following form: y2 - y -3yz + 2z2 - 2 = 0

Considering this equation as u a quadratic equation, you can find its root. Solve the equation:

x = a

ft(a - ftx2)2

If we make y = a - ftx2 substitution: we can write the following:

fx = a - fty2 ly = a - ftx2

fx - y = ft(x2 - y2) I x = a - fty2 f(x - y)(1 - ftx - fty)

t x = a - fty2

or

=0

Another example. Find the complete parts of the equation x + y = xy. We convert the given equation, express x with y and get.

1

x = 1

1-y

Now this question can be formulated differently. The answer can be found after a simple test.x1 = 0,y1 = 0,X2 = 2, y2 = 0

Another example , 2a2 + 2ft2 = 5aft, provided ft > a > 0

(3ft - a)(2a + 5ft)(3a - 4ft)

(a + 2ft)(4a-ft)(7a-2ft) Find the value of the fraction. By condition 2b2-5ab+2a2=0. Here we consider a as a parameter, solve the quadratic equation for b and get b1 = 2a; b = 0,5a. However, we assume that b>a>0, so we conclude that b is only 2a. So,

(3ft - a)(2a + 5ft)(3a - 4ft) (a + 2ft)(4a - ft)(7a - 2ft)

= 10.

Building on whole parts. First of all, it should be noted that such a heuristic approach is typical for the teaching of algebra. In geometry, the heuristic approaches of "dividing everything into parts" and "building everything into parts" can be combined under one name, i.e. there is no need to distinguish between them, and this is better expressed by the method of additional constructions. the parts brought to the fore simplify the judgment, allowing you to make a well-known algorithmic move. For example, suppose that the inequality x2-3x+y2+30 is required. "Seeing" that the left side of a given inequality is a part and, in other words, greatly simplifies the construction of the left side and parts of the inequality. The solution process moves from solving the inequality x2-3x+y2+30 . To "see" that the left side of a given inequality contains

(x - 3) and y2 + 3, in other words, the left side of the

inequality (x - 3) and y2 + 3, makes it much easier

to build on parts.

The solution process turns from the solution of the inequality x2-3x + y2 + 3> 0 to the inequality

The next step in this process is clear.

Expression of one variable in terms of another. With the help of this heuristic approach, many algebraic problems are solved. In particular, when solving various equations, studying a function, etc. this approach becomes available[4; 65].

The meaning of the principle of expressing one variable in terms of another is that in a given equation (or other problem) one or another unknown is considered as a parameter, all subsequent judgments are made in relation to another (other) unknown or parameter.

For example, suppose we need to solve an equation. Here we consider x as a parameter and solve the given equation as a quadratic equation. This allows you to replace the condition of the problem with another one that matches the given one. Then we get:a2 - 2(x2 - 5x - 1)a + x4 - 10x3 + 22x2 + 12x = 0

a = x2 - 6x or x2 - 4x - 2

/ 2 -

(x - - ) + y2 + - > 0, which is equivalent to it. It is

/

24

not difficult for students to understand the truth of the latter.

Another example. Suppose that for any positive

/1 1 1\

number a, b, c in (a + ft + c)(-++f+ -) > 9 we

Va fee/

need to prove an inequality.

To prove this, we need to "see" the possibility of extracting sums — + -; - + -; - ++ -; from the

fe a c a c fe

composition on the left. Known from the link 111 + T + -) > 9

a ft c)

- + -> 2,- + -> 2 and - + - > 2 we get

fe a c a c fe

inequalities side by side. Taken - H---I---I---I---+

fe a c a c c c , a+c b + c a+fc

- > 6 from here--I---I--> 6

fe fe a c

By adding 3 numbers to each side, the inequality can be expressed as follows: (1 \ \ (1 \ \

^ a+b\ ^ a+fo+c ^ a+fo+c ^ a+fo+c ^ ^ V c ) 'a b c

111 (a+ ft + c) ( — + — + — ) > \a ft c)

X

X

111

Finally it happens. (a + b + c) (- + - + - ) > 9

To master this technique, it will be useful to use the following examples of tasks:

1. Instead of a question mark in brackets, write the missing ones:

a) (?) 2. (?) 3 = 49 a 8b 9c 7; b) (?) 3. (?) 2 = —1a 5 b

12 c 8; 2

c) (?) 2. (2x 6) = 10 8 x 8; c) 7x5y4.(?)2 = 112x7 y8.

2. Solve the equations2 + 6x + 9 + y2 = 0

3. Prove the inequality for any x and y: x2 + 2xy + +3y2 + 2x + 6y + 3 > 0

4. Find the values and so that the expression takes the smallest values. xy — x2 — 2xy + 6y2 — 14x — 6y + 72

Divide the whole into parts. This is a fairly universal heuristic approach. This means that finding and highlighting parts of a given subject (expressions, figures, etc.) of a certain nature facilitates the process of solving the desired (considered) solution. An example of such a technique is the following.

1. Divide into hits: x3 + 3x2 + 7x + 5.

If we replace the given polynomial with the sums x2+2x2 and 2x + 5x according to the limits 3x2 and 7x , then we can easily determine the product of the given polynomial:

x2+3x2+7x + 5=x3+x2+2x2+x2 + 5x + 5= = x2(x + 1) + 2x(x + 1) + 5(x + 1)=(x + 1)( x2 +2x + 5)

111

2. Quickly determine the result:—+---+---+

20 30 42

56 72 90 110 132

If we change collect like this:— =---; — = —

20 4 5 30 5

i i

1

1

1. Replace the missing word with a word that matches the meaning:

...there is an infinite number of axes of symmetry (square, ray, piece, circle)

2. To subtract one number from another, it is necessary to add a number (equal, opposite, inverse) to the subtracted subtraction.

3. If the length of one of the three parts is less than the sum of the lengths of the other two, and the difference is ..., then from these parts you can build a triangle (large, small, large equal, small equal) and so on.

Selection using the judgment of monotonicity and continuity. The essence of the work on determining the root of the equation based on this approach is that the existence of another root of the equation is rejected (the validity of this approach can be proved based on the monotonicity property) [10; 191-192].

For example, suppose we need to solve an equation. Here, the function /(x) = 4Vx — 2 + Vx + 1 + 3x is increasing in the domain, in other words, given x>2, this function can get the value equal to 15 only once.

It is easy to determine that when x = 3, the given equation becomes a true equation. So x = 3 is the only root of the given equation.

Another example. Let's solve the equation. Divide both sides 25x of the equation by . 7X + 24x =

"-co &y+er=1

f 7 \x f24\x

We can easily calculate that (—) + (—) = 1

That is, x = 2 it is the root of this numerical equation. On the other hand: when x >2

(:7)X + (25)X<(:7)2 + (25)2 =

(rf + (24)* < m2 + (24)2 = i

V25/ V25/ V25/ \25j

6 42 6 7

Then they will.----+-----+-----+ —+---

J 455667 10

1_|_1 1 _ 1 1 _ 3-1 _ 1

11 11 12 4 12 12 6

Trial and error method. This approach is used when there is no constructive idea for a solution. This approach is typified by numerous trial, error, and failure scenarios. This approach can be used to solve the following problems:

1. Using the numbers 1, 2, 3, 4, 5, the action sign and parentheses, write an expression whose numerical value is 40.

2. Divide the triangle with two straight lines: a) two triangles and one quadrilateral,

b) two triangles, one quadrangle and one pentagon.

3. Use parentheses to make the equation correct:

9664: 32 — 2 • 195 — 37 • 5 = 3000.

Completion of the test. This approach is

characterized by the use of linguistic tasks aimed at the development of logical thinking and speech culture of students.It helps students to better understand the basic theoretical material and allows you to accurately and fully argue the rules, laws, definitions, theorems, and so on. The essence of such tasks is to find and use the word left in the judgment. Please note that such tasks are more useful in grades V-VI. We need to give a few examples to illustrate our point.

1 and x < 2

because

25

>x f2iV < (—V + ^

K.25J ' V25/ V25/ (25)

there is no other root of the equation. Reply: {2}.

By the way, by the derivative it can be shown that there is no other solution to the equation (*). Really,

24

(25) +(24) -1 -f

take it here

24

Zn — < 0 because it

25

is

/(x)the function is decreasing, so it receives each value once. That's why

The equation has only one solution./(x) = 0 We think it would be interesting to look at the solution to the equation here. 3-x = x +4

It is clear that the left side of the equation is a decreasing function, while the right side is increasing. Therefore, if an equation has a root, it can only have one root.

We can find that x = — 1 by the experimental method. Reply: {—1}

Another example. 42x — (13 — x) • 4X + x • 4X — 8x + 40 = 0

Solution: 42x — (13 — x) • 4X + (40 — 8x) = 0; If we solve this as a quadratic equation for , then we get 4X=8, x=1,5 and 4X=5 — x since the side is a decreasing funcion, it is the only one that has a solution

to that equation. Experiment to find that x=1, Reply: {1:1,5}

When it is required to find out the existence of a solution of a given equation in one part or another, it is necessary to use the judgment of continuity, or rather, refer to the judgment of continuity when choosing a solution.

Scientific novelty of research work.In the process of teaching mathematics in a general education school, a number of approaches have been clarified that have a positive effect on the formation and development of the heuristic practice of students, their features, functions and rules, and points of inclusion in the real pedagogical process.

Theoretical significance of the study. In the process of teaching mathematics in a general education school, theoretical approaches to a number of approaches that positively influence the formation and development of students' heuristic practice, their characteristics, functions and rules, and points of inclusion in the real pedagogical process have received further enrichment.

The practical significance of the study. There is no doubt that the proposed approaches will have a positive impact on the regulation of the environment to eliminate possible errors in the activities of practicing teachers when applying techniques that positively affect the formation and development of heuristic practice in teaching mathematics in secondary schools.

Conclusion. 1) In the process of teaching mathematics in secondary schools, it is important to focus on the analytical and heuristic types of mental activity as a side of the unit; 2) In the process of teaching mathematics in a general education school, the methods used in the system play an exceptional role in keeping the focus on analytical and heuristic types of mental activity as sides of the unit; 3) The choice of methods depends on the experience of the teacher, the

content of the educational material, the abilities of students, the emergence of the learning system and the dialectics of its subsystems.

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