What is OGE and its meaning? How to solve tasks in gia

Is it difficult to hand over OGE in mathematics? This question sets himself, perhaps every graduate class 9. Let's deal with together. The main state exam in mathematics is one of the most difficult in grade 9 is a fact. In addition, it is obligatory to pass by each graduate of the main school for receiving the certificate. Therefore, all the difficulties of OGE 2018 in mathematics should be ready in advance.

We want to draw your attention to the fact that you will find qualified tutors for the preparation for the OGE in mathematics for students, and. We practice individual and collective classes of 3-4 people, we provide discounts for training. Our students are on average are gaining 30 points!

To begin with, it is worth noting the first feature of OGE in mathematics, which highlights it among all examination tests not only in 9, but also in the 11th grade. This is, of course, division into modules: "Algebra", "Geometry", "Real Mathematics". If you do not pass the minimum threshold for each of them, it will negatively affect the overall evaluation for the exam.

That is, without taking the right points for at least one of the modules (we recall that in "Algebra" it is 3 points, in "Geometry" - 2, in "Real Mathematics" - 2), it is possible to obtain an "unsatisfactory" estimate for the entire examination work. Thus, the knowledge of students in all sections of the course of mathematics of the main school are checked. Therefore, it should be given enough time to prepare for each block.

Tasks of the "Geometry" module in OGE

So, traditionally in OGE in mathematics, the greatest percentage of unresolved tasks falls on the "Geometry" module. This phenomenon can be found several reasons.

First, the study of geometry at school is allocated by an average three times less than the lessons of algebra. And the material is essentially perceived and digested more complicated and longer than algebraic.

Secondly, the skills of building and reading the drawings in many children are formed poorly and require additional work at home, which most students are, of course, do not.

As a result, the geometry tasks are often simply ignored by students. In other words, they do not even begin to fulfill them. The advice here is the only one: to pay more time to the tasks of geometry throughout the time of preparation. Do not be lazy: look at the solution of similar tasks on the Internet or ask the teacher, then over time the desired skill solution will be formed and on the exam you will be fully armed.

It is worth saying that there are simply no difficult tasks in OGE in math, perhaps, perhaps only tasks 25, 26, and not always. These numbers can also be learned to decide: several learned techniques for the implementation of additional constructions and solutions algorithms will help cope with such tasks.

Tasks of the module "Algebra" in OGE in mathematics

So, we turn to the "Algebra" module. Starting on the first part, perhaps it does not make any sense, all tasks are performed there on fairly simple algorithms, do not require special smelts, everyone learned to solve each student of a secondary school. There is much more interest represent the tasks of part 2. On them we will dwell in more detail.

Task 21 with a solution to OGE in mathematics. Convert an expression, solve equation, solve the system of equations

Fractional rational or powerful expression. The solution requires attention at each conversion step. Consider an example:

Solve inequality

1____ + __1____ + __1____ < 1 (х-3)(х-4) (х-3)(х-5) х²-9х+20 Решение: Для решения данного неравенства выполним следующее 1. Перенесем единицу в левую часть неравенства. 2. Знаменатель третьей дроби разложим на множители (х-4)(х-5) 3.

Since there is a variable in the denominator, it is necessary to specify the OTZ - the area of \u200b\u200bpermissible values \u200b\u200b- those values \u200b\u200bin which the fraction does not make sense. x ≠ 3; x ≠ 4; x ≠ 5 4. Fold four fractions with different denominators (since an integer can be represented as a fraction with a denominator 1), dominant numerals. We get: (x-5) + (x-4) + (x-3) - (x-3) (x-4) (x-5)< 0 3х-12 - (х-3)(х-4)(х-5) < 0 3(х-4) - (х-3)(х-4)(х-5) < 0 Выносим общий множитель (х-4) за скобку (х-4) 〈3 - (х-3)(х-5)〉 < 0 (х-4) 〈3 - (х² - 8х + 15)〉 < 0 (х-4) (3 - х² + 8х - 15) < 0 Коэффициент при х² отрицательный. Меняем его на противоположный, умножая вторую скобку на (-1). При этом изменится знак неравенства на противоположный. (х - 4) (х² - 8х + 12) > 0 (x - 4) (x - 6) (x - 2)\u003e 0 Now we can solve the inequality by intervals. We celebrate on the numeric axis all the roots that we found in the numerator and all the roots of the OTZ from the denominator.

2________3 _________ 4_________ 5_________ 6___________ - in the record, where the coefficient is always positive, the interval method gives the right to apply the following rule: the right to the right root of the inequality sign is always +! When moving through the root, the sign of inequality changes to the opposite.

In case the root has a reader, (for example, x in the square, in the fourth degree, in the sixth, etc.), as in our example with x \u003d 4, the sign of inequality does not change to the opposite. Hence the answer: (-∞, 2) ∪ (3.4) ∪ (4.5) ∪ (6, + ∞).

At every step, a certain nuance of solutions is visible. But in general, the algorithm is understandable and easy to leverave.

Decision of the task 22 in OGE in mathematics. Text task

It does not have to talk here a lot, the textual tasks of the guys, as a rule, decide. Errors may occur at the stage of compiling the equation by the condition of the problem. To avoid such problems, you should be able to correctly formalize the text task, that is, translate from Russian to mathematical language. For this, a large number of techniques are developed: drawings, diagrams, tables, etc. Methods most often used in schools are building tables in movement and work tasks, and schemes in interest tasks. Severe these methods will not be difficult, just desire to do it.

An example of a task 23 in OGE in mathematics. Building complex graphs of functions, expressions with a parameter

Many students say that the most difficult task of OGE in mathematics is number 23. It is difficult to argue with them, the view of such tasks is usually threatening, but in fact, the solution is reduced to the transformation of a large expression into a compact fraction. Moreover, it is enough to know only the rules for the decomposition of polynomials into multipliers and be attentive when they cut the resulting fractions. Building a graph should not cause difficulties, as a last resort, you can always "squeeze" a schedule on points and understand what the function turned out.

After completed constructions, do not forget to complete the assignment: as a rule, you need to define an unknown parameter (number), which ensures the execution of such conditions as one, two, none, etc. Shared points with a graph of the constructed function. Permanent training will help to gain confidence and solve this task without difficulty.

Thus, it is impossible to say that there is a lot of difficult tasks in the OGE in mathematics. The question is only in proper and timely preparation. Attach efforts, and even the most difficult tasks of OGE in mathematics 2018 will seem hopeless to you! Uz "Homograph" sincerely wishes you good luck in the exams!

The main state exam in mathematics must pass all nine-graders to transfer to the 10th grade and receiving the certificate of the main secondary education. For those who want to continue learning in a classroom with a physico-mathematical bias, this test is especially important - you should score the right number of points.

In 2020, some changes made to the exam content. So, it was decided to exclude the "Real Mathematics" module. This does not mean that the relevant issues are simply removed - they will have to be solved in the sections of the "algebra" and "geometry".

Structure of OGE in mathematics

OGE contains 26 tasks that are distributed to two blocks.

The first part contains 20 questions (base level), 14 of them by algebra and 6 - geometry. For the correct solution of each given 1 point. To respond, you need to write a number, digit or sequence of numbers. The student must show how he owns the main algorithms how well knows the concepts and categories. The correctness of the solutions to the computer method is checked.

The second part is 6 OGE tasks in mathematics (elevated and high level of complexity), 3 questions on algebra and as much on geometry, for each one you can get 2 points. To respond, you will need to provide a written decision. This module matters to form a profile group, there will have to provide detailed descriptions. The check is carried out two independent experts, they also make up the protocol.

The exam is given 235 minutes. It is not so much if you take into account the excitement, which usually accompanies such events.

What in your pocket to bring?

Examples are allowed to use reference benefits with some mathematical formulas. But, they do not need to bring them with them - these books are issued to every student during testing. But supplies such as a ruler, a circulation, a pattern for a drawing can be safely taken to the exam. It is not allowed to use a calculator.

To solve all the tasks of OGE 2020 in mathematics and get a good score, it is worth carefully repeating the entire school curriculum. You can work yourself or with a tutor, but the best results show graduates who used demonstration versions when preparing, conveniently connoissed by objects. The main thing is to understand the logic, it is not automatically remembered to solve the questions, but trying to understand the structure and applying the knowledge learned earlier.

Grade 9 "We recruit points" 21 task

FULL NAME: Yurgenson Veronika Aleksandrovna, MBOU "Stepovskaya Sosh"

Work description:

21 tasks from the second part of the OGE in mathematics Includes the following sections:

1. Equations

2. Algebraic expressions

3. Systems of equations

4. Inequality

5. Inequality systems

The tasks of the second part of the algebra module are aimed at checking the ownership of such qualities of mathematical training of graduates as:

formal operational algebraic apparatus;

the ability to solve a comprehensive task that includes knowledge from various those course algebra;

the skills mathematically correctly and clearly record the solution, while leading the necessary explanations and justifications;

ownership with a wide range of receptions and ways of reasoning.

Basic requirements for mathematical preparation

To be able to transform algebraic expressions, solve equations, inequalities and their systems

Sections content elements

Algebraic expressions;

Equations and inequalities

Sections of elements of requirements :

To be able to transform algebraic expressions.

Consider equations which are solved by decomposition on multipliers.

COP code 2; 3.

Code software 2; 3

(x - 2) ² (x-3) \u003d 12 (x-2)

1) (x - 2) ² (x-3) -12 (x-2) \u003d 0

2) (x - 2) ((x - 2) (x-3) -12) \u003d 0

3) (x - 2) (x²-5x-6) \u003d 0

4) x-2 \u003d 0 and x²-5x-6 \u003d 0

5) x \u003d 2; x \u003d -1; x \u003d 6.

Algorithm

We carry out a general factor for brackets (x-2)

We carry out transformations in brackets

Each multiplier equal to zero

We solve the equations, find the roots

2) Consider biquadrate equations that are solved by introducing a new variable

(X-1) 4 -2 (x-1) 2 -3=0

Replacement: (x - 1) ² \u003d T

t²-2t-3 \u003d 0

t \u003d 3 and T \u003d -1

(x - 1) ² \u003d 3 and (x - 1) ² \u003d -1

x²-2x-2 \u003d 0 and x²-2x + 2 \u003d 0

Algorithm

1) We enter a new variable (x - 1) ² \u003dt. ,

2) we get a square equation

3) we solve the square equation, find the roots

4) We return to paragraph 1 replacement

5) solve square equations, find the roots

3) Consider equations that are solved by extracting the root

x² \u003d 6x-5

x²-6x + 5 \u003d 0

x \u003d 1 and x \u003d 5

Algorithm

Remove the root, in this example cubic

We carry all the numbers into the left part, the sign change to the opposite and equal to zero

Solve the obtained equation, find the roots of the equation

COP code 2

CT code 2

The tasks of this type are completely simple if you know the rules for working with degrees - that is, the degree properties

1. Reduce the fraction:

To solve an example of this type, you need to decompose the foundations of degrees to "bricks" - to find such numbers that would be present in the numerator, and in the denominator, and present everything in the form of degrees of these numbers. In this case, these are numbers 2 and 3: ,.

Then:

Answer: 12.

2. Reduce the fraction:

Decision:

Answer: 200.

3. Reduce the fraction:

Decision:

Answer: 33.

Now we will analyze the task in which the degrees are in the letter form:

4. Reduce the fraction:

Decision:

Answer: 0.1 (necessarily through the comma)

5. Reduce the fraction:

In this example, it is possible to solve everything to the degree of twos and to the degree of four:

Decision:

Answer: 0.25

6. Reduce the fraction:

First we transform sums and differences in degrees:

Decision:

Answer: 0.08.

Systems of equations solved by substitution method

COP code 3

CT code 3

Algorithm

1) In the first equation, we will express the variable from x

2) Substitute y \u003d 5-3x in the second equation of the system, we obtain the equation regarding x

3) solve the obtained equation, find the root

4) We substitute x \u003d 3 to the equation y \u003d 5-3x, we find

5) write in response a couple of numbers x and y

Systems of equations solved by algebraic addition

1) 2xqm + 6x \u003d -4

2) 2xqm + 6x + 4 \u003d 0

x \u003d -1 and x \u003d -2

3) 2U² \u003d 8

4) y \u003d -2 and y \u003d 2

5) (-1;-2); (-1;2); (-2;-2); (-2;2)

Algorithm

Mix two system equations

Solumed the resulting square equation

Submount from the first equation the second

Resolved the resulting equation

Write in response a pair of numbers x and

Fractional rational inequalities.

COP code 3

CT code 3

Fractional rational inequalities have the form P (x) / Q (x)\u003e 0 and p (x) / q (x)<0, где P(x),Q(x)-многочлены.

The inequality is equivalent to the following p (x) · Q (x)\u003e 0 and p (x) · q (x)<0, где P(x),Q(x)-многочлены.

The left part of inequality is a whole rational function. The polynomials p (x) and q (x) are decomposed on the factors and solve inequality intervals.

Algorithm

1) Specify the denominator for multipliers

3) answer (because in the inequality sign less in response to write intervals with "-"

Whole rational algebraic inequalities

Such inequalities may be square or linear. Square inequalities are solved somewhat differently by calculating the discriminant. These inequalities, although they have a second degree, but they are solved by bringing to linear, that is, the way of decomposition on linear multipliers. The considered method is called the interval method. The scheme of the solution is as follows.

X \u003d 7 and

Algorithm

1) Transfer to everything into the left of the inequality

2) I will solve this inequality by decomposition by multipliers

3) Now put the points on the straight and define the signs of expressions on each obtained gap

4) Answer (because in the inequality sign less in response to write intervals with "-"

Solve inequality

Decision.

Purchase two parts of inequality in one part and get rid of the denominator: we equate left part to zero and find the roots.

From here and

Preparing the roots on the coordinate direct, we define the signs of inequality, we get: and

Answer: (-∞; -0.75] u)