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The main types of inequalities are presented, including Bernoulli, Cauchy - Bunyakovsky, Minkowski, Chebyshev inequalities. The properties of inequalities and actions on them are considered. The basic methods for solving inequalities are given.
Formulas for basic inequalities
Formulas for universal inequalities
Universal inequalities are satisfied for any values of the quantities included in them. The main types are listed below universal inequalities.
1) | a b | ≤ |a| + |b| ; | a 1 a 2 ... a n | ≤ |a 1 | + |a 2 | + ... + |a n |
2) |a| + |b| ≥ | a - b | ≥ | |a| - |b| |
3)
Equality occurs only when a 1 = a 2 = ... = a n.
4)
Cauchy-Bunyakovsky inequality
Equality holds if and only if α a k = β b k for all k = 1, 2, ..., n and some α, β, |α| + |β| > 0 .
5)
Minkowski's inequality, for p ≥ 1
Formulas of satisfiable inequalities
Satisfiable inequalities are satisfied for certain values of the quantities included in them.
1)
Bernoulli's inequality:
.
In more general view:
,
where , numbers of the same sign and greater than -1
:
.
Bernoulli's Lemma:
.
See "Proofs of inequalities and Bernoulli's lemma".
2)
for a i ≥ 0 (i = 1, 2, ..., n) .
3)
Chebyshev's inequality
at 0 < a 1 ≤ a 2 ≤ ... ≤ a n
And 0 < b 1 ≤ b 2 ≤ ... ≤ b n
.
At 0 < a 1 ≤ a 2 ≤ ... ≤ a n
And b 1 ≥ b 2 ≥ ... ≥ b n > 0
.
4)
Generalized Chebyshev inequalities
at 0 < a 1 ≤ a 2 ≤ ... ≤ a n
And 0 < b 1 ≤ b 2 ≤ ... ≤ b n
and k natural
.
At 0 < a 1 ≤ a 2 ≤ ... ≤ a n
And b 1 ≥ b 2 ≥ ... ≥ b n > 0
.
Properties of inequalities
Properties of inequalities are a set of rules that are satisfied when transforming them. Below are the properties of the inequalities. It is understood that the original inequalities are satisfied for values of x i (i = 1, 2, 3, 4) belonging to some predetermined interval.
1) When the order of the sides changes, the inequality sign changes to the opposite.
If x 1< x 2
,
то x 2 >x 1 .
If x 1 ≤ x 2, then x 2 ≥ x 1.
If x 1 ≥ x 2, then x 2 ≤ x 1.
If x 1 > x 2 then x 2< x 1
.
2) One equality is equivalent to two weak inequalities different sign.
If x 1 = x 2, then x 1 ≤ x 2 and x 1 ≥ x 2.
If x 1 ≤ x 2 and x 1 ≥ x 2, then x 1 = x 2.
3) Transitivity property
If x 1< x 2
и x 2 < x 3
,
то x 1 < x 3
.
If x 1< x 2
и x 2 ≤ x 3
,
то x 1 < x 3
.
If x 1 ≤ x 2 and x 2< x 3
,
то x 1 < x 3
.
If x 1 ≤ x 2 and x 2 ≤ x 3, then x 1 ≤ x 3.
4) The same number can be added (subtracted) to both sides of the inequality.
If x 1< x 2
,
то x 1 + A < x 2 + A
.
If x 1 ≤ x 2, then x 1 + A ≤ x 2 + A.
If x 1 ≥ x 2, then x 1 + A ≥ x 2 + A.
If x 1 > x 2, then x 1 + A > x 2 + A.
5) If there are two or more inequalities with the sign of the same direction, then their left and right sides can be added.
If x 1< x 2
,
x 3 < x 4
,
то x 1 + x 3 < x 2 + x 4
.
If x 1< x 2
,
x 3 ≤ x 4
,
то x 1 + x 3 < x 2 + x 4
.
If x 1 ≤ x 2 , x 3< x 4
,
то x 1 + x 3 < x 2 + x 4
.
If x 1 ≤ x 2, x 3 ≤ x 4, then x 1 + x 3 ≤ x 2 + x 4.
Similar expressions apply to the signs ≥, >.
If the original inequalities contain signs of non-strict inequalities and at least one strict inequality(but all signs have the same direction), then when added, a strict inequality is obtained.
6) Both sides of the inequality can be multiplied (divided) by a positive number.
If x 1< x 2
и A >0, then A x 1< A · x 2
.
If x 1 ≤ x 2 and A > 0, then A x 1 ≤ A x 2.
If x 1 ≥ x 2 and A > 0, then A x 1 ≥ A x 2.
If x 1 > x 2 and A > 0, then A · x 1 > A · x 2.
7) Both sides of the inequality can be multiplied (divided) by a negative number. In this case, the sign of inequality will change to the opposite.
If x 1< x 2
и A < 0
,
то A · x 1 >A x 2.
If x 1 ≤ x 2 and A< 0
,
то A · x 1 ≥ A · x 2
.
If x 1 ≥ x 2 and A< 0
,
то A · x 1 ≤ A · x 2
.
If x 1 > x 2 and A< 0
,
то A · x 1 < A · x 2
.
8) If there are two or more inequalities with positive terms, with the sign of the same direction, then their left and right sides can be multiplied by each other.
If x 1< x 2
,
x 3 < x 4
,
x 1 , x 2 , x 3 , x 4 >0 then x 1 x 3< x 2 · x 4
.
If x 1< x 2
,
x 3 ≤ x 4
,
x 1 , x 2 , x 3 , x 4 >0 then x 1 x 3< x 2 · x 4
.
If x 1 ≤ x 2 , x 3< x 4
,
x 1 , x 2 , x 3 , x 4 >0 then x 1 x 3< x 2 · x 4
.
If x 1 ≤ x 2, x 3 ≤ x 4, x 1, x 2, x 3, x 4 > 0 then x 1 x 3 ≤ x 2 x 4.
Similar expressions apply to the signs ≥, >.
If the original inequalities contain signs of non-strict inequalities and at least one strict inequality (but all signs have the same direction), then multiplication results in a strict inequality.
9) Let f(x) be a monotonically increasing function. That is, for any x 1 > x 2, f(x 1) > f(x 2). Then this function can be applied to both sides of the inequality, which will not change the sign of the inequality.
If x 1< x 2
,
то f(x 1) < f(x 2)
.
If x 1 ≤ x 2 then f(x 1) ≤ f(x 2) .
If x 1 ≥ x 2 then f(x 1) ≥ f(x 2) .
If x 1 > x 2, then f(x 1) > f(x 2).
10) Let f(x) be a monotonically decreasing function, That is, for any x 1 > x 2, f(x 1)< f(x 2)
.
Тогда к обеим частям неравенства можно применить эту функцию, от чего знак неравенства изменится на противоположный.
If x 1< x 2
,
то f(x 1) >f(x 2) .
If x 1 ≤ x 2 then f(x 1) ≥ f(x 2) .
If x 1 ≥ x 2 then f(x 1) ≤ f(x 2) .
If x 1 > x 2 then f(x 1)< f(x 2)
.
Methods for solving inequalities
Solving inequalities using the interval method
The interval method is applicable if the inequality includes one variable, which we denote as x, and it has the form:
f(x) > 0
where f(x) - continuous function, having final number break points. The inequality sign can be anything: >, ≥,<, ≤
.
The interval method is as follows.
1) Find the domain of definition of the function f(x) and mark it with intervals on the number axis.
2) Find the discontinuity points of the function f(x). For example, if this is a fraction, then we find the points at which the denominator becomes zero. We mark these points on the number axis.
3) Solve the equation
f(x) = 0 .
We mark the roots of this equation on the number axis.
4) As a result, the number axis will be divided into intervals (segments) by points. Within each interval included in the domain of definition, we select any point and at this point we calculate the value of the function. If this value is greater than zero, then we place a “+” sign above the segment (interval). If this value is less than zero, then we put a “-” sign above the segment (interval).
5) If the inequality has the form: f(x) > 0, then select intervals with the “+” sign. The solution to the inequality is to combine these intervals, which do not include their boundaries.
If the inequality has the form: f(x) ≥ 0, then to the solution we add points at which f(x) = 0. That is, some intervals may have closed boundaries (the boundary belongs to the interval). the other part may have open boundaries (the boundary does not belong to the interval).
Similarly, if the inequality has the form: f(x)< 0
,
то выбираем интервалы с знаком „-“
.
Решением неравенства будет объединение этих интервалов, в которые не входят их границы.
If the inequality has the form: f(x) ≤ 0, then to the solution we add points at which f(x) = 0.
Solving inequalities using their properties
This method is applicable to inequalities of any complexity. It consists in applying the properties (presented above) to bring the inequalities to more simple view and get a solution. It is quite possible that this will result in not just one, but a system of inequalities. This is a universal method. It applies to any inequalities.
Used literature:
I.N. Bronstein, K.A. Semendyaev, Handbook of mathematics for engineers and college students, “Lan”, 2009.
In mathematics infinite set functions. And each has its own character.) To work with a wide variety of functions you need single approach. Otherwise, what kind of mathematics is this?!) And there is such an approach!
When working with any function, we present it with standard set questions. And the first, the most important question- This domain of definition of the function. This area is sometimes called a set acceptable values argument, function specification area, etc.
What is the domain of a function? How to find it? These questions often seem complex and incomprehensible... Although, in fact, everything is extremely simple. You can see for yourself by reading this page. Let's go?)
Well, what can I say... Just respect.) Yes! The natural domain of a function (which is discussed here) matches With ODZ expressions included in the function. Accordingly, they are searched according to the same rules.
Now let’s look at a not entirely natural domain of definition.)
Additional restrictions on the scope of a function.
Here we will talk about the restrictions that are imposed by the task. Those. the task contains some additional conditions, which were invented by the compiler. Or the restrictions emerge from the very method of defining the function.
As for the restrictions in the task, everything is simple. Usually, there is no need to look for anything, everything is already said in the task. Let me remind you that the restrictions written by the author of the task do not cancel fundamental limitations of mathematics. You just need to remember to take into account the conditions of the task.
For example, this task:
Find the domain of a function:
on the set of positive numbers.
We found the natural domain of definition of this function above. This area:
D(f)=( -∞ ; -1) ∪ (-1; 2] ∪ ∪
IN verbal way When specifying a function, you need to carefully read the condition and find restrictions on X there. Sometimes the eyes look for formulas, but the words whistle past the consciousness yes...) Example from the previous lesson:
The function is specified by the condition: each value of the natural argument x is associated with the sum of the digits that make up the value of x.
It should be noted here that we are talking only O natural values X. Then D(f) instantly recorded:
D(f): x ∈ N
As you can see, the scope of a function is not so complex concept. Finding this region comes down to examining the function, writing a system of inequalities, and solving this system. Of course, there are all kinds of systems, simple and complex. But...
I'll open it little secret. Sometimes a function for which you need to find the domain of definition looks simply intimidating. I want to turn pale and cry.) But as soon as I write down the system of inequalities... And, suddenly, the system turns out to be elementary! Moreover, often, the more terrible the function, the simpler the system...
Moral: the eyes fear, the head decides!)
Scientific supervisor:
1. Introduction 3
2. Historical sketch 4
3. “Place” of ODZ when solving equations and inequalities 5-6
4. Features and dangers of ODZ 7
5. ODZ – there is a solution 8-9
6. Finding ODZ is extra work. Equivalence of transitions 10-14
7. ODZ in the Unified State Exam 15-16
8. Conclusion 17
9. Literature 18
1. Introduction
Problem: equations and inequalities in which it is necessary to find ODZ have not found a place in the algebra course for systematic presentation, which is probably why my peers and I often make mistakes when solving such examples, spending a lot of time solving them, while forgetting about ODZ.
Target: be able to analyze the situation and draw logically correct conclusions in examples where it is necessary to take into account DL.
Tasks:
1. Study theoretical material;
2. Solve many equations, inequalities: a) fractional-rational; b) irrational; c) logarithmic; d) containing inverse trigonometric functions;
3. Apply the studied materials in a situation that differs from the standard one;
4. Create a work on the topic “Area of acceptable values: theory and practice”
Work on the project: I started working on the project by repeating the functions I knew. The scope of many of them is limited.
ODZ occurs:
1. When deciding fractional rational equations and inequalities
2. When deciding irrational equations and inequalities
3. When deciding logarithmic equations and inequalities
4. When solving equations and inequalities containing inverse trigonometric functions
Having solved many examples from various sources(unified state examination manuals, textbooks, reference books), I systematized the solution of examples according to following principles:
· you can solve the example and take into account the ODZ (the most common method)
· it is possible to solve the example without taking into account the ODZ
· it is only possible to come to the right decision by taking into account the ODZ.
Methods used in the work: 1) analysis; 2) statistical analysis; 3) deduction; 4) classification; 5) forecasting.
Studied the analysis Unified State Exam results over the past years. Many mistakes were made in the examples in which DL needs to be taken into account. This once again emphasizes relevance my topic.
2. Historical sketch
Like other concepts of mathematics, the concept of a function did not develop immediately, but went through a long path of development. P. Fermat’s work “Introduction and Study of Flat and Solid Places” (1636, published 1679) says: “Whenever in final equation There are two unknown quantities, there is a place.” Essentially, we are talking about functional dependence and its graphic representation(“place” in Fermat means line). The study of lines according to their equations in R. Descartes’ “Geometry” (1637) also indicates a clear understanding of the mutual dependence of the two variables. In I. Barrow (“Lectures on Geometry”, 1670) in geometric shape the mutual inverse nature of the actions of differentiation and integration is established (of course, without using these terms themselves). This already indicates a completely clear mastery of the concept of function. In geometric and mechanical form We also find this concept in I. Newton. However, the term “function” first appears only in 1692 with G. Leibniz and, moreover, not quite in its modern understanding. G. Leibniz calls various segments associated with a curve (for example, the abscissa of its points) a function. In the first printed course, “Analysis of infinitesimals for the knowledge of curved lines” by L'Hopital (1696), the term “function” is not used.
The first definition of a function in a sense close to the modern one is found in I. Bernoulli (1718): “A function is a quantity composed of a variable and a constant.” This not entirely clear definition is based on the idea of specifying a function analytical formula. The same idea appears in the definition of L. Euler, given by him in his “Introduction to the Analysis of Infinites” (1748): “The function variable quantity is an analytical expression composed in some way from this variable quantity and numbers or constant quantities" However, L. Euler is no longer alien to modern understanding function, which does not connect the concept of a function with any analytical expression of it. In his " Differential calculus” (1755) says: “When certain quantities depend on others in such a way that when the latter change they themselves are subject to change, then the first are called functions of the second.”
WITH early XIX centuries, more and more often they define the concept of a function without mentioning its analytical representation. In "Treatise on Differential and integral calculus"(1797-1802) S. Lacroix says: “Every quantity whose value depends on one or many other quantities is called a function of these latter.” IN " Analytical theory heat" by J. Fourier (1822) there is a phrase: "Function f(x) denotes a completely arbitrary function, that is, a sequence of given values, subordinate or not general law and corresponding to all values x contained between 0 and some value x" The definition of N. I. Lobachevsky is close to modern: “... General concept function requires that the function from x name the number that is given for each x and together with x gradually changes. The function value can be given or analytical expression, or a condition that provides a means of testing all numbers and choosing one of them, or, finally, a dependence may exist and remain unknown.” It is also said there a little lower: “The broad view of the theory allows for the existence of dependence only in the sense that numbers one with another in connection are understood as if given together.” Thus, modern definition functions, free from references to analytical task, usually attributed to P. Dirichlet (1837), was repeatedly proposed before him.
The domain of definition (admissible values) of a function y is the set of values of the independent variable x for which this function is defined, i.e., the domain of change of the independent variable (argument).
3. “Place” of the range of acceptable values when solving equations and inequalities
1. When solving fractional rational equations and inequalities the denominator must not be zero.
2. Solving irrational equations and inequalities.
2.1..gif" width="212" height="51"> .
IN in this case there is no need to find the ODZ: from the first equation it follows that the obtained values of x satisfy the following inequality: https://pandia.ru/text/78/083/images/image004_33.gif" width="107" height="27 src=" > is the system:
Since they enter into the equation equally, then instead of inequality, you can include inequality https://pandia.ru/text/78/083/images/image009_18.gif" width="220" height="49"> 
https://pandia.ru/text/78/083/images/image014_11.gif" width="239" height="51"> 
3. Solving logarithmic equations and inequalities.
3.1. Scheme for solving a logarithmic equation
But it is enough to check only one condition of the ODZ.
3.2..gif" width="115" height="48 src=">.gif" width="115" height="48 src=">
4. Trigonometric equations kind are equivalent to the system (instead of inequality, you can include inequality in the system https://pandia.ru/text/78/083/images/image024_5.gif" width="377" height="23"> are equivalent to the equation
4. Features and dangers of the range of permissible values
In mathematics lessons, we are required to find the DL in each example. At the same time mathematical essence In this case, finding the ODZ is not at all mandatory, often not necessary, and sometimes impossible - and all this without any damage to the solution of the example. On the other hand, it often happens that after solving an example, schoolchildren forget to take into account the DL, write it down as the final answer, and take into account only some conditions. This circumstance is well known, but the “war” continues every year and, it seems, will continue for a long time.
Consider, for example, the following inequality:
Here, the ODZ is sought and the inequality is solved. However, when solving this inequality, schoolchildren sometimes believe that it is quite possible to do without searching for DL, or more precisely, it is possible to do without the condition
In fact, to obtain the correct answer it is necessary to take into account both the inequality , and .
But, for example, the solution to the equation: https://pandia.ru/text/78/083/images/image032_4.gif" width="79 height=75" height="75">
which is equivalent to working with ODZ. However, in this example, such work is unnecessary - it is enough to check the fulfillment of only two of these inequalities, and any two.
Let me remind you that any equation (inequality) can be reduced to the form . The ODZ is simply the domain of definition of the function on the left side. The fact that this area must be monitored follows from the definition of the root as a number from the domain of definition of a given function, thereby from the ODZ. Here funny example on this topic..gif" width="20" height="21 src="> has a domain of definition of a set of positive numbers (this, of course, is an agreement to consider a function with, but reasonable), and then -1 is not a root.
5. Range of acceptable values – there is a solution
And finally, in a lot of examples, finding the ODZ allows you to get the answer without bulky layouts, or even verbally.
1. OD3 is an empty set, which means that the original example has no solutions.
1)
2) 3) 
2. B ODZ one or more numbers are found, and a simple substitution quickly determines the roots.
1)
, x=3
2)
Here in the ODZ there is only the number 1, and after substitution it is clear that it is not a root.
3) There are two numbers in the ODZ: 2 and 3, and both are suitable.
4) > In the ODZ there are two numbers 0 and 1, and only 1 is suitable.
ODZ can be used effectively in combination with analysis of the expression itself.
5) 
< ОДЗ: Но в правой части неравенства могут быть только positive numbers, so we leave x=2. Then we substitute 2 into the inequality.
6) 
From the ODZ it follows that, where we have ..gif" width="143" height="24"> From the ODZ we have: . But then and . Since, there are no solutions.
From the ODZ we have: https://pandia.ru/text/78/083/images/image060_0.gif" width="48" height="24">>, which means . Solving the last inequality, we get x<- 4, что не входит в ОДЗ. Поэтому решения нет.
3) ODZ: . Since then
On the other hand, https://pandia.ru/text/78/083/images/image068_0.gif" width="160" height="24">
ODZ:. Consider the equation on the interval [-1; 0).
It fulfills the following inequalities https://pandia.ru/text/78/083/images/image071_0.gif" width="68" height="24 src=">.gif" width="123" height="24 src="> and there are no solutions. With the function and https://pandia.ru/text/78/083/images/image076_0.gif" width="179" height="25">. ODZ: x>2..gif" width="233" height ="45 src="> Let's find the ODZ:
An integer solution is only possible for x=3 and x=5. By checking we find that the root x=3 does not fit, which means the answer is x=5.
6. Finding the range of acceptable values is extra work. Equivalence of transitions.
You can give examples where the situation is clear even without finding DZ.
1. 
Equality is impossible, because when subtracting a larger expression from a smaller one, the result must be a negative number.
2. 
.
The sum of two non-negative functions cannot be negative.
I will also give examples where finding ODZ is difficult, and sometimes simply impossible.
And finally, searches for ODZ are very often just extra work, which you can do without, thereby proving your understanding of what is happening. There are a huge number of examples that can be given here, so I will choose only the most typical ones. The main solution method in this case is equivalent transformations when moving from one equation (inequality, system) to another.
1.
. ODZ is not needed, because, having found those values of x for which x2 = 1, we cannot obtain x = 0.
2. . ODZ is not needed, because we find out when the radical expression is equal to a positive number.
3. . ODZ is not needed for the same reasons as in the previous example.
4. 
ODZ is not needed, because the radical expression is equal to the square of some function, and therefore cannot be negative.
5. 
6. ..gif" width="271" height="51"> To solve, only one restriction for the radical expression is sufficient. In fact, from the written mixed system it follows that the other radical expression is non-negative.
8. DZ is not needed for the same reasons as in the previous example.
9. 
ODZ is not needed, since it is enough for two of the three expressions under the logarithm signs to be positive to ensure the positivity of the third.
10. 
.gif" width="357" height="51"> ODZ is not needed for the same reasons as in the previous example.
It is worth noting, however, that when solving using the method of equivalent transformations, knowledge of the ODZ (and properties of functions) helps.
Here are some examples.
1. . OD3, which implies that the expression on the right side is positive, and it is possible to write an equation equivalent to this one in this form https://pandia.ru/text/78/083/images/image101_0.gif" width="112" height="27 "> ODZ: But then, and when solving this inequality, it is not necessary to consider the case when right side less than 0.
3. . From the ODZ it follows that, and therefore the case when https://pandia.ru/text/78/083/images/image106_0.gif" width="303" height="48"> The transition in general looks like this:
https://pandia.ru/text/78/083/images/image108_0.gif" width="303" height="24">
There are two possible cases: 0
This means that the original inequality is equivalent to the following set of systems of inequalities:
The first system has no solutions, but from the second we obtain: x<-1 – решение неравенства.
Understanding the conditions of equivalence requires knowledge of some subtleties. For example, why are the following equations equivalent:
Or 
And finally, perhaps most importantly. The fact is that equivalence guarantees the correctness of the answer if some transformations of the equation itself are made, but is not used for transformations in only one of the parts. Abbreviations and the use of different formulas in one of the parts are not covered by the equivalence theorems. I have already given some examples of this type. Let's look at some more examples.
1. This decision is natural. On the left side by property logarithmic function let's move on to the expression ..gif" width="111" height="48">
Having solved this system, we get the result (-2 and 2), which, however, is not an answer, since the number -2 is not included in the ODZ. So, do we need to install ODS? Of course not. But since we used a certain property of the logarithmic function in the solution, we are obliged to provide the conditions under which it is satisfied. Such a condition is the positivity of expressions under the logarithm sign..gif" width="65" height="48">.
2. 
..gif" width="143" height="27 src="> numbers are subject to substitution in this way 
. Who wants to do such tedious calculations?.gif" width="12" height="23 src="> add a condition, and you can immediately see that only the number https://pandia.ru/text/78/083/ meets this condition images/image128_0.gif" width="117" height="27 src=">) was demonstrated by 52% of test takers. One of the reasons for such low rates is the fact that many graduates did not select the roots obtained from the equation after squaring it.
3) Consider, for example, the solution to one of the problems C1: “Find all values of x for which the points of the graph of the function 
lie above the corresponding points of the graph of the function ". The task boils down to solving fractional inequality containing logarithmic expression. We know the methods for solving such inequalities. The most common of them is the interval method. However, when using it, test takers make various mistakes. Let's look at the most common mistakes using inequality as an example:
X< 10. Они отмечают, что в первом случае решений нет, а во втором – корнями являются числа –1 и . При этом выпускники не учитывают условие x < 10.
8. Conclusion
To summarize, we can say that there is no universal method for solving equations and inequalities. Every time, if you want to understand what you are doing and not act mechanically, a dilemma arises: what solution should you choose, in particular, should you look for ODZ or not? I think that the experience I have gained will help me solve this dilemma. I will stop making mistakes by learning how to use ODZ correctly. Whether I can do this, time, or rather the Unified State Examination, will tell.
9. Literature
And others. “Algebra and the beginnings of analysis 10-11” problem book and textbook, M.: “Prosveshchenie”, 2002. “Handbook for elementary mathematics" M.: “Nauka”, 1966. Newspaper “Mathematics” No. 46, Newspaper “Mathematics” No. Newspaper “Mathematics” No. “History of mathematics in school grades VII-VIII”. M.: “Enlightenment”, 1982. etc. “The most complete edition of options real tasks Unified State Exam: 2009/FIPI" - M.: "Astrel", 2009. etc. "Unified State Exam. Mathematics. Universal materials for preparing students/FIPI" - M.: "Intelligence Center", 2009. etc. "Algebra and the beginnings of analysis 10-11." M.: “Enlightenment”, 2007. , “Workshop on solving problems school mathematics(algebra workshop).” M.: Education, 1976. “25,000 mathematics lessons.” M.: “Enlightenment”, 1993. “Preparing for the Olympiads in mathematics.” M.: “Exam”, 2006. “Encyclopedia for children “MATHEMATICS”” volume 11, M.: Avanta +; 2002. Materials from the sites www. *****, www. *****.
