THE METHOD OF İNTERPRETİNG THE LANGUAGE OF A GİVEN PROBLEM İN ANOTHER LANGUAGE İN THE PROCESS OF TEACHİNG MATHEMATİCS Текст научной статьи по специальности «Языкознание и литературоведение»

THE METHOD OF ÍNTERPRETÍNG THE LANGUAGE OF A GÍVEN PROBLEM ÍN ANOTHER LANGUAGE ÍN THE PROCESS OF TEACHÍNG MATHEMATÍCS

Ibrahimov F.,

Sheki branch of ADPU doctor of pedagogical sciences, professor

Aliyeva G. Sheki branch of ADPU, teacher DOI: 10.5281/zenodo.6616158

Abstract

The article gives an explanation of the relevance of the research topic. It is emphasized that depending on the improvement of the organization and management of analytical and heuristic activities in general schools, the level of efficiency of teaching mathematics is conditioned. In the work, attention is drawn to the fact that heuristic activity, which is an important element of the mind, is guided by the selection of tasks presented to students and their transformation into a system. Depending on the application of the mentioned means(types of tasks-question, issue, exercises) in this process, the functions carried by it create different methods and principles as forms of movement of the idea. The interpretation of the given problem language in another language has a special place among those forms of intellectual movement, and it has a positive impact on the formation and development of the pupils ' experience of ephristic activity.

Keywords: priyom; heuristic priyom; issue language; textual issues; equally important issue; the quantity sought; roots of the equation; the multiplicity of possible prices of the unknown.

Relevance of the research topic.

The significant impact of the level of implementation of analytical and heuristic activities of students on improving the efficiency of the process of teaching mathematics in secondary schools is undeniable. From the analysis of scientific sources and the experience of practical teachers, such savings are extracted that the question of including the analytical type of mental activity in the pedagogical process from both theoretical and technological points of view is solved more preferably than the heuristic type. Among the reasons for this reality, a special place is occupied by the fact that heuristic activity is closely related to the creative process, and the types of mental activity are not united as sides of unity. Therefore, it is important to study the place and role of prime numbers in the order of tables that implement the solution of the problem of determining the theoretical and technological directions of using heuristic activity in the pedagogical process, including in the process of teaching mathematics (this is the problem of eternity), as well as to study the definition of their content. Based on this consideration, we affirm the relevance of the research on the topic: "a simple way to interpret the language of a problem in another language in the process of teaching mathematics."

Interpretation of research material.

As it is known, the specific activity of the subject with the purpose of learning is training. This process includes: 1) the acquisition of information about the essential properties of the world necessary for the successful organization of this or that type of ideal and practical activity; 2) mastering the ways and operations that make up these activities; 3) to master the rules of using the mentioned information in order to correctly select and control the ways and operations corresponding to the conditions of the issue and the forthcoming purpose. Thus, training is possible where human activity is guided by conscious goals in order to acquire certain knowledge, habits and skills [5; 192].

Training is an important characteristic of educational activity and does not cover all its aspects. Training, in the broadest sense of the word, new knowledge, involves the acquisition of skills and habits. However, mastery and teaching activities are essentially different events. Assimilation is not only the learning process, but an integral part of every field of activity [2; 150]. The structure of teaching activities has the following components: teaching situation (or tasks), teaching operations, control, assessment [7; 9]. Teaching tasks involve students discovering and mastering general solutions to various problems and specific practical issues in their teaching activities. By proposing teaching tasks, the teacher introduces students to a specific teaching situation. There are many specific features of teaching tasks, but one thing is clear: in order to solve them, students must master the necessary teaching operations.

Priyom is an integral part of the method, or a separate side of it, 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 learn the techniques of cognitive activity in a special and systematic way in the process of acquiring knowledge, skills and habits [8; 42-43]. The mastery of these techniques not only has a positive effect on the formation of knowledge, but also changes the attitude of students to learning, making it a purposeful, organized, independent subject [3; 244]. When structures are focused on the organization and management of analytical and heuristic activities, their structure (executive steps) is focused on [5; 119]. The approach presented here is aimed at mastering the experience of heuristic activity.

The peculiarity of this method is that the transition to the same problem occurs here (first of all, from the algorithmic point of view). Thus, the given issue takes a different form in terms of language, it is interpreted in another language. [4; 73].

For example, let's say

fx2 + y2 = 36

I y = X3

is required to determine how many solutions the system has. Change the text of the given problem by drawing geometric means.

We express it as follows: "How many points do the circle with radius 6 cm, the origin of the central coordinate and the cube-parabola y = x3 have?" Undoubtedly, the solution of the latter is obvious.

The solution of textual problems by algebraic methods is, of course, based on the transition to symbolic language. As a result, a system of equations, inequalities, equations or inequalities is required according to the condition of the problem. [4; 73]

Let's give some examples to make what we say clearer and clearer.

l.Issue. The cyclist traveled the first part of the road at a speed of vi, km/h (h, less than m kilometers) and the second part at a speed of v, km / h. Knowing that the average speed of the cyclist is v km/h, find out how much the road goes.

Solution: Let the first part of the road be x km. It took — hours to get it. The second vi section of the road

Vi

(x-m) is km. It took him hours to go. So to go all

v2

the way (^ + --—) the hour time is S = x + (x m) = 2x — m. According to the condition, it takes time 2x m hours to go this way.

Here we get the equation. This is the solution of

,, x , x—m 2x—m

the equation.—\--=-

Vi V2

X = ■

mv1(v — v2)

vv1 + vv2 — 2v1v2 And all the way

2mv1(v — v2)

S = 2x — m =

m

The solution of this equation is still the whole

way

x = -—shows who you

VV1+W2—V1V2

are.

2. Solve the inequality.x > -By translating the given symbolic information into verbal form, we can get the problem: "Find all the numbers, each of which is greater than its inverse." The latter can be resolved by a simple oral hearing.

The possibilities of applying this heuristic

approach are very wide. This method has a special place in the teaching of almost all mathematics courses. [10; 191]

The following examples confirm this.

3.l3x — 2l+x<lx+5l — 1 solve the inequality.

When the unknown is on a real axis with a set of possible values x = 2 expression (3x — 2)and when x = —5, the expression (x + 5)becomes zero.

-s 0

Therefore, the designated area of inequality to

these points

(-:ro; —5), [—5; 2], (~; is divided into three

parts. Inequality based on the above a){ X < 5

a)[—(3x — 2) + x < — (x + 5) — 1

.A —5<x<-

b){ 3 l—(3x — 2) + x<(x + 5) — 1

( x>2-

v){ 3

l(3x — 2)+x<(x + 5) — 1 are brought into the system of three inequalities such as.

Let's solve the received systems:

x <5 ^ ( x < —5

1 ~ u

a){

(x < —5 { x > 8

b)

O { o

3x + 2<—x — 5 — 1 i—x < —8

o x e 0

2

— 5 - X -3 o

3x + 2+ x<x+5 — 1

5-1-2 of-5"-2

—3x <2 I x > —2

vv-i + vv2 —2v1v2 mv(vi —v2)

vvi + vv2 — 2viv2

While solving this issue, another quantity (the first part of the road), which depends on the desired quantities (the whole road), was indicated with a sign.

Now, pointing to the desired quantity of self-knowledge, let's resolve the issue again. If there is an entire road, the first part of the road will be and

the second part will be km. The cyclist goes to the first part of the ^+^clock, the second part to the

2vi

^clock. According to the terms of the case, the cyclist went all the way to the -clock. From here

x+m , x—m x 1 , . 1

--\--= - equation is obtained.

2v1 2v2 V ^

Taken from here.x £

K: -a

v)

x>-

3

3x — 2 + x<x + 4 l3x < 6

o{x> 3 o

x>3 ^xe(2-;2)

x <2 V3 >

The result is a set of points in the solution interval

2

of the given inequality. (- ; 2)

4. Issue. The sum of the three numbers that make up the number series is 30. If we take 5 from the first number, 4 from the second number and do not change the third, the resulting numbers form a geometric series. Find these numbers.

Solution: Let us denote the given numbers by a1,a2,a3. Then ax+a2+a3= 30 and a1 — 5;a2 — 4 ,a3 it is because it forms a geometric series. (a2 — 4)2 = a3 (a- — 5) So we can write the following system.

( a1+ a2+ a3 = 30 l(a2 — 4)2 = a3(ai — 5)

Thus, we translate the verbal relationship between numbers into algebraic (symbolic language). By solving the system

a1 + a1d+ a3+2d = 30 3a1+3d = 30

a1+d = 10, a2 = 10

36 = (a1 + 2d)(a1 — 5) 36 = (10 + d)(10 — d — 5)

36 = (10 + d)(d — 5) 36 = 50 — 10d + 5d — d2 d2 + 5d — 14 = 0 d1 = 2,d2 = —7 a1 = 8,al = 17 8; 10; 12 or 17; 10; We determine that 3. 5. Depending on the representation of some equations, their solution is simplified by expressing the variable in the equation to other variables and bringing the given equation to the system of equations. This can be illustrated by the following example:

For example. Solution:

(13-x)V13-x+(5+x)V5+x _

,_ _ 1

(13-x)iI5+x+(5+x)iI13-x

Find the range of possible values of the variable

(PML).

(13 — x > 0 rî1,n ^ „ ^ x e i-5,131 1-5+^ 0 L ' *_

iI13 — x= U> 0 and iI5 + x _ v > 0 let point.

Then, if we consider the notation in the given equation,

it will look like this

ÎU3+v3 -_ 1

U2v+Uv2

U2v+Uv2

U2 + v2 = 18

The system of equations is obtained. From here U3 +v3

1 « f

\U2v + Uv2 ~ o U2 +v2 = 18

(U2 + Uv + v2 = Uv U2 + v2 = 18

o

(U = v IM = 3

system of equations, more general equations can also be solved in the form lax ±b ± Vex ± b = g. To do this, you need to know the method of solving quadratic and cubic equations (Fresnel and Cardan methods). More common

J a — f(x) + nJb+f(x) = g(x) equations in the form of

U = J a — f(x) and U =JJb + f(x) carried out the replacement

( U + v = g(x)

{Un + vn =a + b can be solved by bringing it to the system. There are also irrational equations that are one of the most convenient ways to solve them by bringing them to the trigonometric equation with the help of trigonometric substitutions.

Let's explain this with an example. 7+ = —solve the equation.

X Jl-x2 12 M

To solve the equation, first find the DMQO. 1 — x2 >0, xe]—1;0[v]0;1[ So let's replace it with x = sint. The equation with the new variable will be as follows:

1 1 35 ■ + ■

sint |cost| 12

1

/hen cost> 0.--+ ■

sint

cost + sint 35

1 1 35

1) we get when cost> 0.--\--= —

sint cost 12

sintcost 12 sint + cost = U If we replace it with, we can

U2-1

[\v\ = 3 is taken.

Since it is U1, v > 0 we get U = V-3 from here. If we write them in the markup, then finally x=4. The test shows that this number is the root of the equation. Answer: {4}

Let's solve a more difficult type of equation with this method. For example,

Va — bx ± \lbx — c = d Consider an example of the equation.

6. V77 + x + V20 — x = 5 Solve the equation: V77 + x, = U, V20 — x = v If we denote by, the given equation becomes the following system of equations equivalent to it:

( U + v = 5 lu4 + v4 =97 In the first equation U2 + v2= 25- 2Uv is obtained, which is also the second equation.

Considering (U2+v2)2 —2U2v2 = 97 if we take into account the converted form, (25 — 2Uv)2 — 2U2v2 = 97 and it means (Uv)2 — 50Uv + 264 = 0.

If we solve this equation, (U + v = 5 ,(U + v = 5 l Uv = 6 and {Uv = 44 As the system of equations is taken. If we solve this system of equations and consider the notation, we can x1 = —61x2 = 4.

Examination shows that these numbers are the roots of the equation. Answer:{—61; 4}

It is appropriate to note here that by reducing to a

sint • cost =

2

2U

Then, taking into account the substitution, =

35 7 5

->U1=-,U2 = — 5

12 1 5 2 7

77

sint + cost = — or sint + cost =--

5 5

7

sint = - - cost Solve the equation by squaring

4 3

both sides of the equation, we get sint = - or sint -.

3 4 5 5

So, x, = - or x2=—

' 1 5 2 5

77 sint + cost = — o cost = — — sint.

If we solve both sides of this equation by squaring,

. -5±V73

sint = -we get. So,

14 b '

-5+173 -5-V73 ,

x3=-, x4=-happens.

3 14 4 14

Obviously, when you raise the square, you need to check.

Therefore, checking the obtained numbers (as a result of verification) we conclude that the numbers are

the root of the equation. x = -,- , 5+V73

5 5 4

1 1 35

2) when cost <0 we get---= —

sint cost 12

If we take the cost-sint=U substitution with the

77

same rule U1 = — - or U2 = - we will get. Then cost — sint = — 5 or cost — sin t = 5.

77

If we solve these equations with the above rule, we will still find the same roots (the equation can be solved by bringing it to the system).

. (-S±V73 3 41

Answer: {-; -; —}

(.14 5/ 5,J

It should be noted that some irrational equations can be solved by studying the functions included in the equation. Let's give an example of this.

8. 15 — x + x — 7 = lx — 1 solve the equation. Obviously, if we square both sides of this equation, we get a four-degree equation that is also difficult to solve. So, first of all, let's find DMGO. (5 — x > 0 _ M C1

il5 — x + x — 7 < il5 — 1 + x — 7 = x —5 <0

since there is a solution to the equation

N5 —x + x—7 = lx—1

{ lx—1 = 0 should be.

There is no solution to this system. So there is no solution to the equation. Note that when solving irrational equations with inequalities, it is also possible to use the study of the function by derivative.

Scientific novelty of research work. In the process of teaching mathematics in secondary schools, a number of features, functions and rules and points of inclusion of the "method of interpreting the language of a given problem in another language" in the real pedagogical process were clarified.

Theoretical significance of the research. In the process of teaching mathematics in secondary schools, theoretical approaches to a number of features, functions and rules and points of incorporation into the real pedagogical process of the "method of interpreting a given problem language in another language", which has a positive impact on the formation and development of students' heuristic practice.

Practical significance of the research. There is no doubt that the proposed approaches will have a positive impact on the regulation of the environment in which practitioners work to eliminate the errors that can be made in the application of the "method of interpreting the language of a given problem in another language", which has a positive impact on the formation and development of students' heuristic practice.

The result.l) In the process of teaching mathematics in secondary schools, it is important to

keep analytical and heuristic types of mental activity in the center of attention as aspects of the unit; 2) In the process of teaching mathematics in secondary schools, the methods have an exceptional role in keeping the analytical and heuristic types of mental activity in the center of attention as aspects of the unit; 3) The choice of any type of reception is conditioned by the teacher's experience, the content of the training material, the potential of the students, the emergent nature of the training system and the dialectic of its subsystems.

References

1. Abbasov AN, Mammadzade RR, Mammadli LA Pedagogy: Muntakhabat. Baku, "Translator", 2021, p. 275-284.

2. Alizade AA Psychological problems of modern Azerbaijani school. Baku, "Pedagogy", 2014, pp.150153.

3. Ibrahimov FN Essays on the basics of optimal ratios of algorithmic and heuristic activity in training. Baku, "Translator", 2020, pp.131-140.

4. Ibrahimov FN Application of tasks according to the main argument in teaching mathematics. Baku, "Translator", 2021, p. 52-80.

5. General psychology (edited by AVPetrovsky). Baku, "Maarif', 1985, p.131-133.

6. Mahmutov MI Theory and practice of problem-based learning. Kazan, Tatar Book Publishing House, 1972, p. 119.

7. Эflbконнн B. Intellectual abilities of young schoolchildren and content of training. On Sat. "Age possibilities of knowledge acquisition, ed. D.B. Elkonina and VV Davydova. Moscow. Enlightenment, 1966, p.9

8. Skatkin MN Activation of cognitive activity of students in training. Moscow, Enlightenment, 1965, pp.42-43.

9. Zhuikov SF The problem of activation of students in the psychology of teaching and education.// "Soviet pedagogy", 1966, №1.

10. Dorofeev GV, Patopov MK, Fozov NH Mathematics manual for graduates of universities. Moscow, Enlightenment, p.191