Preparing for the exam solving simple equations. Preparation for the exam in mathematics of the basic and specialized level

Demo versions USE 2018 years of tasks for solving simple equations are located under the number 7 for basic level and under number 5 for the profile level.

In general form, an equation with one variable can be written as f(x) = g(x) , i.e. as equality, which can (but does not have to) contain a variable in both parts x... For example:

sin x = 0,5; 15 = x 2 − 4x; x + 8 = √ x − 18______ ; log 2 ( x+ 5) = log 0.5 8.

What does "solve the equation" mean? Solving an equation means finding all its roots or proving that they do not exist.

Definition: The root of the equation with one variable, such a value of this variable is called, which turns it into a true numerical equality.

Equations with the same roots are called tantamount to... Solving the equation, we strive to transform it to a simpler form. Do not forget to make sure that the simpler equation remains the same as the original one. How to do this for different types of equations, I will mention as I go through the examples. Common to all equations are the following, well known to you,

rules of equivalent transformations:

If in the equation any term is transferred from one part to another, changing its sign, then an equation equivalent to the given one will be obtained.

If both sides of the equation are multiplied or divided by the same nonzero number, then an equation is obtained that is equivalent to the given one.

Taking into account the specifics of this part of the exam, we conclude that there cannot be equations without roots in this task, because the format of the answer does not allow for the required proof. And if the equation has more than one root, then the condition of the problem will be formulated taking into account this fact. For example, "in your answer indicate the smaller of the roots (the largest, the largest negative, the smallest positive ...)".

And using the concept of a root, we conclude that the answer obtained is easy to check on your own. Therefore, when solving this task of the USE in mathematics, an important stage is verification.

The check should be carried out directly according to the condition of the problem printed on the form. Otherwise, you do not check the possibility that you may have inadvertently rewritten the condition in the draft.

So the answer found is x 0 must be substituted into the condition of the problem and make sure that the expressions on the right and left sides of the equality take equal numerical values.

Example.

Find the root of the equation x − 119 ______ x + 7 = −5.

This equation is of the fractional-rational type. The simplest form of such equations p(x) ___ q(x) = 0.

Because if a certain fraction is equal to zero, then we can reason logically: this is possible in the case when the numerator of the fraction is zero, and the denominator is not zero, because you cannot divide by zero. The latter must be taken into account in order to get rid of possible "false" ("extra") roots.

Solution.

1) We transform the fraction to a simple form.
To do this, let's move everything to the right side (do not forget about changing the sign of the term when passing through the equal sign!). Then we bring the fraction to a common denominator.

x − 119 ______ x + 7 + 5 = 0.

x − 119 ______ x + 7 + x + 7 / 5 = 0.

x − 119 + 5(x + 7) x + 7 _______________ = 0.

x − 119 + 5x + 35 _______________ x + 7 = 0.

6x − 84 ______ x + 7 = 0.

2) Equate the numerator of the fraction to zero:
6x − 84 = 0;
6x = 84;
x = 84/6 = 14.

3) For the denominator, write down the condition
x + 7 ≠ 0.
Next, we choose which is simpler,
- substitute in the last inequality x= 14 to make sure the assumed root found is not "false": x + 7 = 14 + 7 = 21 ≠ 0,
or
- solve the opposite equation, in order to then discard the roots coinciding for the numerator and denominator: x + 7 = 0; x = −7; −7 ≠ 14.
In this case, both approaches are simple and obvious.

4) We draw a conclusion - the value of the variable x= 14, which turns the numerator to zero, and is the desired root of the equation.

Answer: 14.

Before rewriting the answer into the form, we do a check.

Examination.

1) We take the initial condition of the problem.

x − 119 ______ x + 7 = −5.

2) Instead of x substitute our answer: 14.

14 − 119 _______ 14 + 7 = −5.

3) We calculate the numerical values of each part of the equality separately. In this example, there is already a number on the right side, so we only calculate the left side.
14 − 119 = −105; 14 + 7 = 21; −105/21 = −5.

4) Since −5 = −5, then x= 14 is the root of the equation, and you can safely rewrite the answer in the form.

Simple equations in one variable.

All the equations that you solved at school, and which, accordingly, can be found in this USE task in mathematics, can be divided into several basic types - rational, irrational, exponential, logarithmic, trigonometric. You can find the exact definitions of these terms in the tutorial. Here we will be interested only in the classification according to the types of those equations that are presented in the bank of USE assignments for problems with a short answer. It is needed in order to be able to "recognize the equation by sight" and immediately guess where to start solving it.

And you usually need to start solving any equation by converting it to its simplest form. The simplest, as a rule, is such a way of writing the equation, which coincides with the "General view" presented in the textbook. Because it is for this method of writing that there are recommendations for getting an answer. And it is these recommendations that you passed in the lessons, they are set out in the textbooks.

Below you will see a table that will help you navigate the variety of equations offered in this exam in mathematics. In it, the symbol x denotes a variable whose unknown value is to be found. The vast majority of equations use the same notation. However, do not forget that other characters, such as y, z, u, v, t,..., have the right to exist as unknowns, including in equations with one variable. Other symbols in the "General view" column - a, b, c- constants are indicated, i.e. the constants for this notation of the equation are quantities. Simply put, in a specific case, numbers will simply stand in their place.

And finally, notation with brackets - p(x), q(x), f(x), g(x) are expressions. In the classroom, you should have heard the term "mathematical expression" more than once. However, if this still does not mean anything to you, then call it for yourself, for example, the formula from x.

Initially, something in this table may seem confusing to you. Skip this and come back to it again after analyzing the next group of examples, as well as immediately before the exam, to quickly review all the possible options that may be encountered in this task.

Attention: The table is clickable. If you left-click on one of the equations in the third column, the solution of this example will be loaded. But do not rush to do it. First, think about how you yourself will solve it. Then compare the answers. Your solution doesn't have to be the same as mine. The main criterion for correctness is obtaining an identity by substituting a root in the original equation.

Square

Equation type		General form	Examples of tasks	Signs
Rational	Linear	ax = b		Equality contains only numbers and x in the first degree.
	ax 2 + bx + c = 0		Numbers, x and x 2. Presence x 2 is required.
	Rational integers containing a polynomial of degree n> 2	p(x) = 0, Where p(x) - polynomial		Numbers and x in varying degrees. There is a degree greater than 2.
	Fractional rational.	p(x) ___ q(x) = 0.		There is NS in the denominator.
Irrational		n √ f(x)____ = n √ g(x)____ ; N √ f(x)____ = φ (x)

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Equations, part $ C $

Equality containing an unknown number denoted by a letter is called an equation. The expression to the left of the equal sign is called the left side of the equation, and the expression to the right is called the right side of the equation.

Scheme for solving complex equations:

Before solving the equation, it is necessary to write down the range of permissible values (ODV) for it.
Solve the equation.
Choose from the obtained roots of the equation that which satisfy the ODZ.

ODZ of various expressions (by expression we mean an alphanumeric notation):

1. The expression in the denominator must not be zero.

$ (f (x)) / (g (x)); g (x) ≠ 0 $

2. The radical expression must not be negative.

$ √ (g (x)); g (x) ≥ 0 $.

3. The radical expression in the denominator must be positive.

$ (f (x)) / (√ (g (x))); g (x)> 0 $

4. For a logarithm: sub-logarithmic expression must be positive; the basis must be positive; the base cannot be equal to one.

$ log_ (f (x)) g (x) \ table \ (\ g (x)> 0; \ f (x)> 0; \ f (x) ≠ 1; $

Logarithmic Equations

Logarithmic equations are equations of the form $ log_ (a) f (x) = log_ (a) g (x) $, where $ a $ is a positive number other than $ 1 $, and equations that reduce to this form.

To solve logarithmic equations, it is necessary to know the properties of logarithms: we will consider all the properties of logarithms for $ a> 0, a ≠ 1, b> 0, c> 0, m $ - any real number.

1. For any real numbers $ m $ and $ n $ the equalities are true:

$ log_ (a) b ^ m = mlog_ (a) b; $

$ log_ (a ^ m) b = (1) / (m) log_ (a) b. $

$ log_ (a ^ n) b ^ m = (m) / (n) log_ (a) b $

$ log_ (3) 3 ^ (10) = 10log_ (3) 3 = 10; $

$ log_ (5 ^ 3) 7 = (1) / (3) log_ (5) 7; $

$ log_ (3 ^ 7) 4 ^ 5 = (5) / (7) log_ (3) 4; $

2. The logarithm of the product is equal to the sum of the logarithms in the same base of each factor.

$ log_a (bc) = log_ (a) b + log_ (a) c $

3. The logarithm of the quotient is equal to the difference between the logarithms of the numerator and denominator for the same base

$ log_ (a) (b) / (c) = log_ (a) b-log_ (a) c $

4. When multiplying two logarithms, you can swap their bases

$ log_ (a) b ∙ log_ (c) d = log_ (c) b ∙ log_ (a) d $, if $ a, b, c $ and $ d> 0, a ≠ 1, b ≠ 1. $

5. $ c ^ (log_ (a) b) = b ^ (log_ (a) b) $, where $ a, b, c> 0, a ≠ 1 $

6. Formula for the transition to a new base

$ log_ (a) b = (log_ (c) b) / (log_ (c) a) $

7. In particular, if it is necessary to swap the base and the sub-logarithmic expression

$ log_ (a) b = (1) / (log_ (b) a) $

There are several main types of logarithmic equations:

Simplest logarithmic equations: $ log_ (a) x = b $. The solution of this type of equations follows from the definition of the logarithm, i.e. $ x = a ^ b $ and $ x> 0 $

We represent both sides of the equation in the form of the logarithm to the base $ 2 $

$ log_ (2) x = log_ (2) 2 ^ 3 $

If the logarithms are equal in the same base, then the sub-logarithmic expressions are also equal.

Answer: $ x = 8 $

Equations of the form: $ log_ (a) f (x) = log_ (a) g (x) $. Because the grounds are the same, then we equate the sub-logarithmic expressions and take into account the ODZ:

$ \ table \ (\ f (x) = g (x); \ f (x)> 0; \ g (x)> 0, а> 0, а ≠ 1; $

$ log_ (3) (x ^ 2-3x-5) = log_ (3) (7-2x) $

Because the bases are the same, then we equate sublogarithmic expressions

We transfer all terms to the left-hand side of the equation and present similar terms

Let us check the found roots by the conditions $ \ table \ (\ x ^ 2-3x-5> 0; \ 7-2x> 0; $

When substituted into the second inequality, the root $ x = 4 $ does not satisfy the condition, therefore, it is an extraneous root

Answer: $ x = -3 $

Variable replacement method.

In this method, you need:

Write down the ODZ equation.
According to the properties of logarithms, achieve the same logarithms in the equation.
Replace $ log_ (a) f (x) $ with any variable.
Solve the equation for the new variable.
Return to step 3, substitute the value for the variable and get the simplest equation of the form: $ log_ (a) x = b $
Solve the simplest equation.
After finding the roots of the logarithmic equation, it is necessary to put them in item 1 and check the condition of the ODZ.

Solve the equation $ log_ (2) √x + 2log_ (√x) 2-3 = 0 $

1. Let us write down the ODZ equation:

$ \ table \ (\ x> 0, \ text "since it stands under the sign of the root and the logarithm"; \ √x ≠ 1 → x ≠ 1; $

2. Let's make the logarithms to the base $ 2 $, for this we use the rule of transition to a new base in the second term:

$ log_ (2) √x + (2) / (log_ (2) √x) -3 = 0 $

4. We obtain a fractional - rational equation with respect to the variable t

Let us reduce all terms to a common denominator $ t $.

$ (t ^ 2 + 2-3t) / (t) = 0 $

The fraction is zero when the numerator is zero and the denominator is nonzero.

$ t ^ 2 + 2-3t = 0 $, $ t ≠ 0 $

5. Let's solve the resulting quadratic equation by Vieta's theorem:

6. Let's return to item 3, make the reverse change and obtain two simple logarithmic equations:

$ log_ (2) √x = 1 $, $ log_ (2) √x = 2 $

Let us logarithm the right-hand sides of the equations

$ log_ (2) √x = log_ (2) 2 $, $ log_ (2) √x = log_ (2) 4 $

Equating sub-logarithmic expressions

$ √x = 2 $, $ √x = 4 $

To get rid of the root, square both sides of the equation

$ x_1 = 4 $, $ x_2 = 16 $

7. Substitute the roots of the logarithmic equation in item 1 and check the condition of the AED.

$ \ (\ table \ 4> 0; \ 4 ≠ 1; $

The first root satisfies the ODV.

$ \ (\ table \ 16> 0; \ 16 ≠ 1; $ The second root also satisfies ODV.

Answer: $ 4; 16 $

Equations of the form $ log_ (a ^ 2) x + log_ (a) x + c = 0 $. Such equations are solved by introducing a new variable and passing to the usual quadratic equation. After the roots of the equation are found, it is necessary to select them taking into account the ODV.

Fractional Rational Equations

If the fraction is zero, then the numerator is zero and the denominator is non-zero.
If at least one part of a rational equation contains a fraction, then the equation is called fractional rational.

To solve a fractionally rational equation, you need:

Find the values of the variable for which the equation does not make sense (ODV)
Find the common denominator of the fractions in the equation;
Multiply both sides of the equation by a common denominator;
Solve the resulting whole equation;
To exclude from its roots those that do not satisfy the DHS condition.

If the equation involves two fractions and their numerators are equal expressions, then the denominators can be equated to each other and the resulting equation can be solved without paying attention to the numerators. BUT given the ODV of the entire original equation.

Exponential Equations

Equations in which the unknown is contained in the exponent are called exponential.

When solving exponential equations, the properties of degrees are used, recall some of them:

1. When multiplying degrees with the same bases, the base remains the same, and the indicators are added.

$ a ^ n a ^ m = a ^ (n + m) $

2. When dividing degrees with the same bases, the base remains the same, and the indicators are subtracted

$ a ^ n: a ^ m = a ^ (n-m) $

3. When raising a power to a power, the base remains the same, and the indicators are multiplied

$ (a ^ n) ^ m = a ^ (n ∙ m) $

4. When raising to the power of a product, each factor is raised to this power

$ (a b) ^ n = a ^ n b ^ n $

5. When raising to a power of a fraction, the numerator and denominator are raised to this power

$ ((a) / (b)) ^ n = (a ^ n) / (b ^ n) $

6. When raising any base to a zero exponent, the result is one

7. The base in any negative exponent can be represented as a base in the same positive exponent by changing the position of the base relative to the fraction line

$ a ^ (- n) = (1) / (a ^ n) $

$ (a ^ (- n)) / (b ^ (- k)) = (b ^ k) / (a ^ n) $

8. The radical (root) can be represented as a power with a fractional exponent

$ √ ^ n (a ^ k) = a ^ ((k) / (n)) $

Types of exponential equations:

1. Simple exponential equations:

a) The form $ a ^ (f (x)) = a ^ (g (x)) $, where $ a> 0, a ≠ 1, x $ is unknown. To solve such equations, we use the property of degrees: degrees with the same base ($ a> 0, a ≠ 1 $) are equal only if their exponents are equal.

b) Equation of the form $ a ^ (f (x)) = b, b> 0 $

To solve such equations, both sides must be logarithm to the base $ a $, it turns out

$ log_ (a) a ^ (f (x)) = log_ (a) b $

2. Method of equalizing the bases.

3. Method of factorization and change of variable.

For this method, in the entire equation by the property of degrees, it is necessary to transform the degrees to one form $ a ^ (f (x)) $.
Change the variable $ a ^ (f (x)) = t, t> 0 $.
We get a rational equation, which must be solved by factoring the expression.
We do the reverse replacement, taking into account that $ t>

Solve the equation $ 2 ^ (3x) -7 2 ^ (2x-1) + 7 2 ^ (x-1) -1 = 0 $

By the property of degrees, we transform the expression so that we get the degree 2 ^ x.

$ (2 ^ x) ^ 3- (7 (2 ^ x) ^ 2) / (2) + (7 2 ^ x) / (2-1) = 0 $

Let's change the variable $ 2 ^ x = t; t> 0 $

We obtain a cubic equation of the form

$ t ^ 3- (7 t ^ 2) / (2) + (7 t) / (2) -1 = 0 $

Multiply the whole equation by $ 2 $ to get rid of the denominators

$ 2t ^ 3-7 t ^ 2 + 7 t-2 = 0 $

Expand the left side of the equation by the grouping method

$ (2t ^ 3-2) - (7 t ^ 2-7 t) = 0 $

Take out from the first bracket the common factor $ 2 $, from the second $ 7t $

$ 2 (t ^ 3-1) -7t (t-1) = 0 $

Additionally, in the first parenthesis we see the formula difference of cubes

$ (t-1) (2t ^ 2 + 2t + 2-7t) = 0 $

The product is zero when at least one of the factors is zero

1) $ (t-1) = 0; $ 2) $ 2t ^ 2 + 2t + 2-7t = 0 $

Let's solve the first equation

Let us solve the second equation in terms of the discriminant

$ D = 25-4 2 2 = 9 = 3 ^ 2 $

$ t_2 = (5-3) / (4) = (1) / (2) $

$ t_3 = (5 + 3) / (4) = 2 $

$ 2 ^ x = 1; 2 ^ x = (1) / (2); 2 ^ x = 2 $

$ 2 ^ x = 2 ^ 0; 2 ^ x = 2 ^ (- 1); 2 ^ x = 2 ^ 1 $

$ x_1 = 0; x_2 = -1; x_3 = 1 $

Answer: $ -1; 0; 1 $

4. Quadratic conversion method

We have an equation of the form $ A a ^ (2f (x)) + B a ^ (f (x)) + C = 0 $, where $ A, B $ and $ C $ are coefficients.
We make the substitution $ a ^ (f (x)) = t, t> 0 $.
A quadratic equation of the form $ A t ^ 2 + B t + C = 0 $ is obtained. We solve the resulting equation.
We do the reverse replacement, taking into account the fact that $ t> 0 $. We get the simplest exponential equation $ a ^ (f (x)) = t $, solve it and write the result in the answer.

Factoring methods:

Factor out the common factor.

To factor a polynomial by factoring out the common factor outside the parentheses, you need:

Determine the common factor.
Divide the given polynomial by it.
Write the product of the common factor and the resulting quotient (enclosing this quotient in parentheses).

Factor the polynomial: $ 10a ^ (3) b-8a ^ (2) b ^ 2 + 2a $.

The common factor for this polynomial is $ 2a $, since all terms are divisible by $ 2 $ and by "a". Next, we find the quotient of dividing the original polynomial by "2a", we get:

$ 10a ^ (3) b-8a ^ (2) b ^ 2 + 2a = 2a ((10a ^ (3) b) / (2a) - (8a ^ (2) b ^ 2) / (2a) + ( 2a) / (2a)) = 2a (5a ^ (2) b-4ab ^ 2 + 1) $

This is the end result of the factorization.

Applying abbreviated multiplication formulas

1. The square of the sum is decomposed into the square of the first number plus twice the product of the first number by the second number and plus the square of the second number.

$ (a + b) ^ 2 = a ^ 2 + 2ab + b ^ 2 $

2. The square of the difference is decomposed into the square of the first number minus twice the product of the first number by the second and plus the square of the second number.

$ (a-b) ^ 2 = a ^ 2-2ab + b ^ 2 $

3. The difference of the squares is decomposed into the product of the difference of numbers and their sum.

$ a ^ 2-b ^ 2 = (a + b) (a-b) $

4. The cube of the sum is equal to the cube of the first number plus three times the square of the first number by the second number plus three times the product of the first and the square of the second number plus the cube of the second number.

$ (a + b) ^ 3 = a ^ 3 + 3a ^ 2b + 3ab ^ 2 + b ^ 3 $

5. The cube of the difference is equal to the cube of the first number minus the triple product of the square of the first number by the second number, plus the triple product of the first and the square of the second number, and minus the cube of the second number.

$ (a-b) ^ 3 = a ^ 3-3a ^ 2b + 3ab ^ 2-b ^ 3 $

6. The sum of the cubes is equal to the product of the sum of numbers by the incomplete square of the difference.

$ a ^ 3 + b ^ 3 = (a + b) (a ^ 2-ab + b ^ 2) $

7. The difference of the cubes is equal to the product of the difference of numbers by the incomplete square of the sum.

$ a ^ 3-b ^ 3 = (a-b) (a ^ 2 + ab + b ^ 2) $

Grouping method

The grouping method is convenient to use when it is necessary to factorize a polynomial with an even number of terms. In this method, it is necessary to collect the terms in groups and take out the common factor from each group outside the bracket. For several groups, after placing in brackets, we should get the same expressions, then we move this bracket as a common factor forward and multiply by the bracket of the resulting quotient.

Factor polynomial $ 2a ^ 3-a ^ 2 + 4a-2 $

To expand this polynomial, we use the method of grouping the terms, for this we group the first two and the last two terms, while it is important to put the sign in front of the second grouping correctly, we put the + sign and therefore write the terms with their signs in brackets.

$ (2a ^ 3-a ^ 2) + (4a-2) = a ^ 2 (2a-1) +2 (2a-1) $

After taking out the common factors, we got a pair of identical parentheses. Now we take out this bracket as a common factor.

$ a ^ 2 (2a-1) +2 (2a-1) = (2a-1) (a ^ 2 + 2) $

The product of these parentheses is the end result of the factorization.

Using the square trinomial formula.

If there is a square trinomial of the form $ ax ^ 2 + bx + c $, then it can be expanded by the formula

$ ax ^ 2 + bx + c = a (x-x_1) (x-x_2) $, where $ x_1 $ and $ x_2 $ are the roots of the square trinomial

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EQUATIONS IN THE USE ON MATH EXAMPLES AND SOLUTIONS Kravchenko N.А. Mathematics teacher GBOU SOSH №891, Moscow Educational presentation to prepare for the Unified State Exam

CONTENTS Task Annotation Example 1 (irrational equation) Example 2 (exponential equation) Example 3 (irrational equation) Example 4 (fractional rational equation) Example 5 (logarithmic equation) Example 6 (logarithmic equation) Example 7 (trigonometric equation) Example 8 ( exponential equation) Example 9 (irrational equation) Example 10 (logarithmic equation)

TYPE OF REFERENCE: Equation. CHARACTERISTIC OF THE PROBLEM: Simple exponential, logarithmic, trigonometric or irrational equation. COMMENT: The equation is reduced in one action to linear or square (in this case, the answer needs to indicate only one of the roots - greater or less). Incorrect answers are mainly due to arithmetic errors.

Solve the equation. EXAMPLE 1 Solution. Let's square it: Then we get where the Answer is: -2

EXAMPLE 2 Solve the equation. Solution. Let's move on to one base of the degree: From equality of bases to equality of degrees: Whence Answer: 3

EXAMPLE 3 Solve the equation. Solution. Let's raise both sides of the equation to the third power: After elementary transformations we get: Answer: 23

EXAMPLE 4 Solve the equation. If your equation has more than one root, indicate the smaller one in your answer. Solution. Range of valid values: x ≠ 10. On this area, we multiply by the denominator: Both roots lie in the ODZ. The lesser of them is −3. Answer: -3

EXAMPLE 5 Solve the equation. Solution. Using the formula we get: Answer: 6

EXAMPLE 6 Solve the equation. Solution. The logarithms of two expressions are equal if the expressions themselves are equal and at the same time positive: From where we get the Answer: 6

EXAMPLE 7 Solve the equation. Indicate the smallest positive root in your answer. Solution. Let's solve the equation:

Large positive roots correspond to the values. If k = 1, then x 1 = 6.5 and x 2 = 8.5. If k = 0, then x 3 = 0.5 and x 4 = 2.5. The values correspond to the smaller values of the roots. The smallest positive decision is 0.5. Answer: 0, 5

EXAMPLE 8 Solve the equation. Solution. Reducing the left and right sides of the equation to powers of 6, we get: Whence it means, Answer: 2

EXAMPLE 9 Solve the equation. Solution. Squaring both sides of the equation, we get: Obviously from where Answer: 5

EXAMPLE 10 Solve the equation. Solution. Let's rewrite the equation so that on both sides there is a logarithm to the base 4: Further, it is obvious where the Answer comes from: -11

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The development and selection of tasks for the formation of knowledge, skills and abilities is a very important task. To achieve this goal, two types of problems are used - purely mathematical and practice-oriented. Day ...

The video course "Get an A" includes all the topics necessary to successfully pass the exam in mathematics at 60-65 points. Completely all tasks 1-13 of the Profile Unified State Exam in Mathematics. Also suitable for passing the Basic exam in mathematics. If you want to pass the exam for 90-100 points, you need to solve part 1 in 30 minutes and without mistakes!

Preparation course for the exam for grades 10-11, as well as for teachers. Everything you need to solve part 1 of the exam in mathematics (first 12 problems) and problem 13 (trigonometry). And this is more than 70 points on the exam, and neither a hundred-point student nor a humanities student can do without them.

All the theory you need. Quick solutions, traps and secrets of the exam. Disassembled all the relevant tasks of part 1 from the Bank of tasks of the FIPI. The course fully meets the requirements of the exam-2018.

The course contains 5 large topics, 2.5 hours each. Each topic is given from scratch, simple and straightforward.

Hundreds of exam assignments. Word problems and probability theory. Simple and easy to remember algorithms for solving problems. Geometry. Theory, reference material, analysis of all types of USE assignments. Stereometry. Tricky solutions, helpful cheat sheets, spatial imagination development. Trigonometry from scratch to problem 13. Understanding instead of cramming. Visual explanation of complex concepts. Algebra. Roots, degrees and logarithms, function and derivative. The basis for solving complex problems of the 2nd part of the exam.