There is not much time left for ninth grade graduates before they have to take the main state exam. This is a very important stage in life, since many students will go to study at technical schools and colleges, and in order to enter the desired budget place you need to do well on your tests. I will solve the OGE grade 9 - just an indispensable site. It will help you prepare for testing much faster than with self-study, in order to pass it with the highest mark of "5".

How to prepare for exams?

In order to prepare for exams, students use different methods. This applies to the study of additional literature, classes with a professional tutor, as well as additional lessons with a school teacher.

Still the most effective method is considered, of course, the use of specialized sites, such as "I will solve the OGE." It helps to prepare both children from the fifth and 9th grade.

Reshu OGE website

Why is this service so popular? It makes it possible to feel the same as in the case of the exam itself. For preparation, tests are given from previous years, because according to statistics, most of the “new” tasks will be very similar to those in previous years.

An important advantage is that you do not need to solve tickets in a complex every time, if this is not necessary. You can perform separate tasks on a specific topic, which will be very convenient if you need to prepare for specific knowledge.

How to find the necessary information on the site?

What does any visitor see as soon as he enters the portal? At the very top of the page is the site header, and below it, in convenient icons, are the names of those subjects that you can choose for the exam. First of all, there are the following:

• mathematics;
• physics;
• chemistry;
• Russian language;
• Informatics.

Disciplines

This list is not complete, because in order to find the necessary subject for which you need to prepare, you just need to go to the site. You can immediately select the desired discipline and then the portal will show all the information on this subject.

Under the list of items are fifteen popular tickets selected by the moderators as indicative.

Test Options

If the student passes only them, and then analyzes his mistakes together with the teacher, then this will increase his chances of success by several times. successful solution OGE for grade 9.

Option No. 6561231

New User Registration

Such a desire to solve the OGE for grade 9 is natural for any student. This requires good preparation. To use the entire service with already completed tasks in full, you must go through the registration process. This will make it possible not only to pass as many tests as you want, but also to keep your statistics.

Statistics in personal account

It will allow you to understand what tasks you need to work on in order to significantly raise the level of knowledge to the required level. You can also share this data with a teacher or tutor so that they can determine what topics are best to draw the student's attention to and what to work on further.

Registration data

To register for the site Reshu OGE Grade 9, it is important to indicate certain user data, including the following:

• address Email;
• password;
• teacher or student.

The most important thing in this case will be to specify the email. Since useful information for the user will begin to come to the registered address. Additionally, it is worth noting the possibility that if the student forgets his password, then using e-mail it will be possible to restore this information. This means that a new temporary code will be sent to the address, which can then be replaced.

Catalog of popular tasks

Job catalog

After the user has successfully registered on the website Reshu OGE Grade 9, namely, the students of this class will be fully prepared for the exams. In the list on the left, you can find a button that says "Task Catalog" and then click on it.

There, all the tasks are already divided by topic, and you can safely go to the place with which information you need to work out further. For example, select "Actions with ordinary fractions". By clicking on this link, the student will get acquainted with the list of tasks that he may have on the exam.

Useful information for experts

School of Experts

This site is visited not only by students, but also by teachers, who will subsequently be engaged in checking assignments. Because each form must be checked in the same way as hundreds of thousands of others without prejudice to the student.

To learn more about the information, it is important to go to the "Expert" tab. There are specific guidelines to check each job. Also, for training, you can start checking specifically selected tasks, and then get comments on grading: how to do it right, and how to avoid mistakes next time.

The unique site "I will solve the OGE" will help you prepare more effectively for the main state exam. Each student will know exactly what to expect on the test, and all examiners will be familiar with the requirements for checking papers.

DECIDE OGE is an application for preparing for the state. exam. Thanks to him, you can train to complete tasks right on your Android device.

Important information

This solution does not offer a theoretical part. To study the theory, you should use other programs or resort to school manuals.

Exam preparation

RESHU OGE gives you the opportunity to prepare for the upcoming test online or offline - at the student's choice. Among the available subjects are the Russian language, mathematics, computer science, geography, history, physics and chemistry. You can prepare in several modes, including the exam simulation mode, in which the user is very limited in time.

By solving the examples and tasks offered by the program, you can turn on the view of the correct answers at the end of the entire test or check with the answers after passing each of the tasks. Don't forget that in order to run a test offline, you first need to download it to your Android device's memory.

Usage

The application is aimed at repeating previously studied material in preparation for passing the OGE. Forget textbooks that are hard to carry around in your bag all the time! Forget that you can only prepare at home or in classroom, conveniently laying out notes and reference books on the desk. With the RESHU OGE application, you can train in solving tasks anytime and anywhere.

Key features

• offers a solution to tasks for training before the main state exam;
• does not contain a theoretical part;
• works without a network connection;
• can simulate a real exam (limit the user in time);
• has a simple and intuitive interface;
• works on all current versions of the Android operating system;
• distributed free of charge.

Secondary general education

Line UMK G.K. Muravina. Algebra and the beginnings of mathematical analysis (10-11) (deep)

Line UMK Merzlyak. Algebra and the Beginnings of Analysis (10-11) (U)

Mathematics

Preparation for the exam in mathematics ( profile level): tasks, solutions and explanations

We analyze tasks and solve examples with the teacher

Examination paper profile level lasts 3 hours 55 minutes (235 minutes).

Minimum Threshold- 27 points.

The examination paper consists of two parts, which differ in content, complexity and number of tasks.

The defining feature of each part of the work is the form of tasks:

• part 1 contains 8 tasks (tasks 1-8) with a short answer in the form of an integer or a final decimal fraction;
• part 2 contains 4 tasks (tasks 9-12) with a short answer in the form of an integer or a final decimal fraction and 7 tasks (tasks 13-19) with a detailed answer (full record of the decision with the rationale for the actions performed).

Panova Svetlana Anatolievna, mathematic teacher the highest category schools, 20 years of work experience:

“In order to get a school certificate, a graduate must pass two mandatory exams in the form of the Unified State Examination, one of which is mathematics. In accordance with the Concept for the Development of Mathematical Education in Russian Federation The USE in mathematics is divided into two levels: basic and specialized. Today we will consider options for the profile level.

Task number 1- checks the ability of USE participants to apply the skills acquired in the course of 5-9 grades in elementary mathematics in practical activities. The participant must have computational skills, be able to work with rational numbers, be able to round decimals be able to convert one unit of measurement to another.

Example 1 In the apartment where Petr lives, a cold water meter (meter) was installed. On the first of May, the meter showed an consumption of 172 cubic meters. m of water, and on the first of June - 177 cubic meters. m. What amount should Peter pay for cold water for May, if the price of 1 cu. m of cold water is 34 rubles 17 kopecks? Give your answer in rubles.

Solution:

1) Find the amount of water spent per month:

177 - 172 = 5 (cu m)

2) Find how much money will be paid for the spent water:

34.17 5 = 170.85 (rub)

Answer: 170,85.

Task number 2- is one of the simplest tasks of the exam. The majority of graduates successfully cope with it, which indicates the possession of the definition of the concept of function. Task type No. 2 according to the requirements codifier is a task for using the acquired knowledge and skills in practical activities and everyday life. Task No. 2 consists of describing, using functions, various real relationships between quantities and interpreting their graphs. Task number 2 tests the ability to extract information presented in tables, diagrams, graphs. Graduates need to be able to determine the value of a function by the value of the argument with various ways of specifying the function and describe the behavior and properties of the function according to its graph. It is also necessary to be able to find the maximum or smallest value and build graphs of the studied functions. The mistakes made are of a random nature in reading the conditions of the problem, reading the diagram.

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Example 2 The figure shows the change in the exchange value of one share of a mining company in the first half of April 2017. On April 7, the businessman purchased 1,000 shares of this company. On April 10, he sold three-quarters of the purchased shares, and on April 13 he sold all the remaining ones. How much did the businessman lose as a result of these operations?

Solution:

2) 1000 3/4 = 750 (shares) - make up 3/4 of all purchased shares.

6) 247500 + 77500 = 325000 (rubles) - the businessman received after the sale of 1000 shares.

7) 340,000 - 325,000 = 15,000 (rubles) - the businessman lost as a result of all operations.

Answer: 15000.

Task number 3- is a task basic level the first part, tests the ability to perform actions with geometric shapes on the content of the course "Planimetry". Task 3 tests the ability to calculate the area of ​​a figure on checkered paper, the ability to calculate degree measures corners, calculate perimeters, etc.

Example 3 Find the area of ​​a rectangle drawn on checkered paper with a cell size of 1 cm by 1 cm (see figure). Give your answer in square centimeters.

Solution: To calculate the area of ​​this figure, you can use the Peak formula:

To calculate the area given rectangle Let's use Pick's formula:

S= B +	G
	2

where V = 10, G = 6, therefore

S = 18 +	6
	2

Answer: 20.

See also: Unified State Examination in Physics: solving vibration problems

Task number 4- the task of the course "Probability Theory and Statistics". The ability to calculate the probability of an event in the simplest situation is tested.

Example 4 There are 5 red and 1 blue dots on the circle. Determine which polygons are larger: those with all red vertices, or those with one of the blue vertices. In your answer, indicate how many more of one than the other.

Solution: 1) We use the formula for the number of combinations from n elements by k:

all of whose vertices are red.

3) One pentagon with all red vertices.

4) 10 + 5 + 1 = 16 polygons with all red vertices.

whose vertices are red or with one blue vertex.

whose vertices are red or with one blue vertex.

8) One hexagon whose vertices are red with one blue vertex.

9) 20 + 15 + 6 + 1 = 42 polygons that have all red vertices or one blue vertex.

10) 42 - 16 = 26 polygons that use the blue dot.

11) 26 - 16 = 10 polygons - how many polygons, in which one of the vertices is a blue dot, are more than polygons, in which all vertices are only red.

Answer: 10.

Task number 5- the basic level of the first part tests the ability to solve the simplest equations (irrational, exponential, trigonometric, logarithmic).

Example 5 Solve Equation 2 3 + x= 0.4 5 3 + x .

Solution. Divide both sides of this equation by 5 3 + X≠ 0, we get

2 3 + x	= 0.4 or		2		3 + X	=	2	,
5 3 + X			5				5

whence it follows that 3 + x = 1, x = –2.

Answer: –2.

Task number 6 in planimetry for finding geometric quantities (lengths, angles, areas), modeling real situations in the language of geometry. The study of the constructed models using geometric concepts and theorems. The source of difficulties is, as a rule, ignorance or incorrect application of the necessary theorems of planimetry.

Area of ​​a triangle ABC equals 129. DE- median line parallel to side AB. Find the area of ​​the trapezoid ABED.

Solution. Triangle CDE similar to a triangle CAB at two corners, since the corner at the vertex C general, angle CDE equal to the angle CAB as the corresponding angles at DE || AB secant AC. Because DE is the middle line of the triangle by the condition, then by the property of the middle line | DE = (1/2)AB. So the similarity coefficient is 0.5. The areas of similar figures are related as the square of the similarity coefficient, so

Hence, S ABED = S Δ ABC – S Δ CDE = 129 – 32,25 = 96,75.

Task number 7- checks the application of the derivative to the study of the function. For successful implementation, a meaningful, non-formal possession of the concept of a derivative is necessary.

Example 7 To the graph of the function y = f(x) at the point with the abscissa x 0 a tangent is drawn, which is perpendicular to the straight line passing through the points (4; 3) and (3; -1) of this graph. Find f′( x 0).

Solution. 1) Let's use the equation of a straight line passing through two given points and find the equation of a straight line passing through points (4; 3) and (3; -1).

(y – y 1)(x 2 – x 1) = (x – x 1)(y 2 – y 1)

(y – 3)(3 – 4) = (x – 4)(–1 – 3)

(y – 3)(–1) = (x – 4)(–4)

–y + 3 = –4x+ 16| · (-1)

y – 3 = 4x – 16

y = 4x– 13, where k 1 = 4.

2) Find the slope of the tangent k 2 which is perpendicular to the line y = 4x– 13, where k 1 = 4, according to the formula:

3) The slope of the tangent is the derivative of the function at the point of contact. Means, f′( x 0) = k 2 = –0,25.

Answer: –0,25.

Task number 8- checks the knowledge of elementary stereometry among the exam participants, the ability to apply formulas for finding surface areas and volumes of figures, dihedral angles, compare the volumes of similar figures, be able to perform actions with geometric figures, coordinates and vectors, etc.

The volume of a cube circumscribed around a sphere is 216. Find the radius of the sphere.

Solution. 1) V cube = a 3 (where A is the length of the edge of the cube), so

A 3 = 216

A = 3 √216

2) Since the sphere is inscribed in a cube, it means that the length of the diameter of the sphere is equal to the length of the edge of the cube, therefore d = a, d = 6, d = 2R, R = 6: 2 = 3.

Task number 9- requires the graduate to transform and simplify algebraic expressions. Task number 9 advanced level Difficulty with short answers. Tasks from the section "Calculations and transformations" in the USE are divided into several types:

transformations of numerical rational expressions;

transformations of algebraic expressions and fractions;

transformations of numerical/letter irrational expressions;

actions with degrees;

transformation logarithmic expressions;

1. conversion of numeric/letter trigonometric expressions.

Example 9 Calculate tgα if it is known that cos2α = 0.6 and

3π	< α < π.
4

Solution. 1) Let's use the double argument formula: cos2α = 2 cos 2 α - 1 and find

tan 2 α =	1	– 1 =	1	– 1 =	10	– 1 =	5	– 1 = 1	1	– 1 =	1	= 0,25.
	cos 2 α		0,8		8		4		4		4

Hence, tan 2 α = ± 0.5.

3) By condition

3π	< α < π,
4

hence α is the angle of the second quarter and tgα< 0, поэтому tgα = –0,5.

Answer: –0,5.

#ADVERTISING_INSERT# Task number 10- checks the ability of students to use the acquired early knowledge and skills in practical activities and everyday life. We can say that these are problems in physics, and not in mathematics, but all the necessary formulas and quantities are given in the condition. The tasks are reduced to solving a linear or quadratic equation, or a linear or quadratic inequality. Therefore, it is necessary to be able to solve such equations and inequalities, and determine the answer. The answer must be in the form of a whole number or a final decimal fraction.

Two bodies of mass m= 2 kg each, moving at the same speed v= 10 m/s at an angle of 2α to each other. The energy (in joules) released during their absolutely inelastic collision is determined by the expression Q = mv 2 sin 2 α. At what smallest angle 2α (in degrees) must the bodies move so that at least 50 joules are released as a result of the collision?
Solution. To solve the problem, we need to solve the inequality Q ≥ 50, on the interval 2α ∈ (0°; 180°).

mv 2 sin 2 α ≥ 50

2 10 2 sin 2 α ≥ 50

200 sin2α ≥ 50

Since α ∈ (0°; 90°), we will only solve

We represent the solution of the inequality graphically:

Since by assumption α ∈ (0°; 90°), it means that 30° ≤ α< 90°. Получили, что наименьший угол α равен 30°, тогда наименьший угол 2α = 60°.

Task number 11- is typical, but it turns out to be difficult for students. The main source of difficulties is the construction of a mathematical model (drawing up an equation). Task number 11 tests the ability to solve word problems.

Example 11. During spring break, 11-grader Vasya had to solve 560 training problems to prepare for the exam. On March 18, on the last day of school, Vasya solved 5 problems. Then every day he solved the same number of problems more than the previous day. Determine how many problems Vasya solved on April 2 on the last day of vacation.

Solution: Denote a 1 = 5 - the number of tasks that Vasya solved on March 18, d– daily number of tasks solved by Vasya, n= 16 - the number of days from March 18 to April 2 inclusive, S 16 = 560 - the total number of tasks, a 16 - the number of tasks that Vasya solved on April 2. Knowing that every day Vasya solved the same number of tasks more than the previous day, then you can use the formulas for finding the sum arithmetic progression:

560 = (5 + a 16) 8,

5 + a 16 = 560: 8,

5 + a 16 = 70,

a 16 = 70 – 5

a 16 = 65.

Answer: 65.

Task number 12- check students' ability to perform actions with functions, be able to apply the derivative to the study of the function.

Find the maximum point of a function y= 10ln( x + 9) – 10x + 1.

Solution: 1) Find the domain of the function: x + 9 > 0, x> –9, that is, x ∈ (–9; ∞).

2) Find the derivative of the function:

4) The found point belongs to the interval (–9; ∞). We define the signs of the derivative of the function and depict the behavior of the function in the figure:

The desired maximum point x = –8.

Download for free the work program in mathematics to the line of UMK G.K. Muravina, K.S. Muravina, O.V. Muravina 10-11 Download free algebra manuals

Task number 13- an increased level of complexity with a detailed answer, which tests the ability to solve equations, the most successfully solved among tasks with a detailed answer of an increased level of complexity.

a) Solve the equation 2log 3 2 (2cos x) – 5log 3 (2cos x) + 2 = 0

b) Find all the roots of this equation that belong to the segment.

Solution: a) Let log 3 (2cos x) = t, then 2 t 2 – 5t + 2 = 0,

	log3(2cos x) =	2	⇔		2cos x = 9	⇔		cos x =	4,5	⇔ because |cos x| ≤ 1,
	log3(2cos x) =	1			2cos x = √3			cos x =	√3
		2							2

then cos x =	√3
	2

	x =	π	+ 2π k
		6
	x = –	π	+ 2π k, k ∈ Z
		6

b) Find the roots lying on the segment .

It can be seen from the figure that the given segment has roots

11π	And	13π	.
6		6

Answer: A)	π	+ 2π k; –	π	+ 2π k, k ∈ Z; b)	11π	;	13π	.
	6		6		6		6

Task number 14- advanced level refers to the tasks of the second part with a detailed answer. The task tests the ability to perform actions with geometric shapes. The task contains two items. In the first paragraph, the task must be proved, and in the second paragraph, it must be calculated.

The circumference diameter of the base of the cylinder is 20, the generatrix of the cylinder is 28. The plane intersects its bases along chords of length 12 and 16. The distance between the chords is 2√197.

a) Prove that the centers of the bases of the cylinder lie on the same side of this plane.

b) Find the angle between this plane and the plane of the base of the cylinder.

Solution: a) A chord of length 12 is at a distance = 8 from the center of the base circle, and a chord of length 16, similarly, is at a distance of 6. Therefore, the distance between their projections onto a plane, parallel to the bases cylinders is either 8 + 6 = 14 or 8 − 6 = 2.

Then the distance between chords is either

= = √980 = = 2√245

= = √788 = = 2√197.

According to the condition, the second case was realized, in which the projections of the chords lie on one side of the axis of the cylinder. This means that the axis does not intersect this plane within the cylinder, that is, the bases lie on one side of it. What needed to be proven.

b) Let's denote the centers of the bases as O 1 and O 2. Let us draw from the center of the base with a chord of length 12 the perpendicular bisector to this chord (it has a length of 8, as already noted) and from the center of the other base to another chord. They lie in the same plane β perpendicular to these chords. Let's call the midpoint of the smaller chord B, greater than A, and the projection of A onto the second base H (H ∈ β). Then AB,AH ∈ β and, therefore, AB,AH are perpendicular to the chord, that is, the line of intersection of the base with the given plane.

So the required angle is

∠ABH = arctan	AH	= arctg	28	= arctg14.
	BH		8 – 6

Task number 15- an increased level of complexity with a detailed answer, checks the ability to solve inequalities, the most successfully solved among tasks with a detailed answer of an increased level of complexity.

Example 15 Solve the inequality | x 2 – 3x| log 2 ( x + 1) ≤ 3x – x 2 .

Solution: The domain of definition of this inequality is the interval (–1; +∞). Consider three cases separately:

1) Let x 2 – 3x= 0, i.e. X= 0 or X= 3. In this case, this inequality becomes true, therefore, these values ​​are included in the solution.

2) Let now x 2 – 3x> 0, i.e. x∈ (–1; 0) ∪ (3; +∞). In this case, this inequality can be rewritten in the form ( x 2 – 3x) log 2 ( x + 1) ≤ 3x – x 2 and divide by a positive expression x 2 – 3x. We get log 2 ( x + 1) ≤ –1, x + 1 ≤ 2 –1 , x≤ 0.5 -1 or x≤ -0.5. Taking into account the domain of definition, we have x ∈ (–1; –0,5].

3) Finally, consider x 2 – 3x < 0, при этом x∈ (0; 3). In this case, the original inequality will be rewritten in the form (3 x – x 2) log 2 ( x + 1) ≤ 3x – x 2. After dividing by a positive expression 3 x – x 2 , we get log 2 ( x + 1) ≤ 1, x + 1 ≤ 2, x≤ 1. Taking into account the area, we have x ∈ (0; 1].

Combining the obtained solutions, we obtain x ∈ (–1; –0.5] ∪ ∪ {3}.

Answer: (–1; –0.5] ∪ ∪ {3}.

Task number 16- advanced level refers to the tasks of the second part with a detailed answer. The task tests the ability to perform actions with geometric shapes, coordinates and vectors. The task contains two items. In the first paragraph, the task must be proved, and in the second paragraph, it must be calculated.

In an isosceles triangle ABC with an angle of 120° at the vertex A, a bisector BD is drawn. Rectangle DEFH is inscribed in triangle ABC so that side FH lies on segment BC and vertex E lies on segment AB. a) Prove that FH = 2DH. b) Find the area of ​​the rectangle DEFH if AB = 4.

Solution: A)

1) ΔBEF - rectangular, EF⊥BC, ∠B = (180° - 120°) : 2 = 30°, then EF = BE due to the property of the leg lying opposite the angle of 30°.

2) Let EF = DH = x, then BE = 2 x, BF = x√3 by the Pythagorean theorem.

3) Since ΔABC is isosceles, then ∠B = ∠C = 30˚.

BD is the bisector of ∠B, so ∠ABD = ∠DBC = 15˚.

4) Consider ΔDBH - rectangular, because DH⊥BC.

2x	=	4 – 2x
2x(√3 + 1)		4

1	=	2 – x
√3 + 1		2

√3 – 1 = 2 – x

x = 3 – √3

EF = 3 - √3

2) S DEFH = ED EF = (3 - √3 ) 2(3 - √3 )

S DEFH = 24 - 12√3.

Answer: 24 – 12√3.

Task number 17- a task with a detailed answer, this task tests the application of knowledge and skills in practical activities and everyday life, the ability to build and explore mathematical models. This task - text task with economic content.

Example 17. The deposit in the amount of 20 million rubles is planned to be opened for four years. At the end of each year, the bank increases the deposit by 10% compared to its size at the beginning of the year. In addition, at the beginning of the third and fourth years, the depositor annually replenishes the deposit by X million rubles, where X - whole number. Find highest value X, at which the bank will add less than 17 million rubles to the deposit in four years.

Solution: At the end of the first year, the contribution will be 20 + 20 · 0.1 = 22 million rubles, and at the end of the second - 22 + 22 · 0.1 = 24.2 million rubles. At the beginning of the third year, the contribution (in million rubles) will be (24.2 + X), and at the end - (24.2 + X) + (24,2 + X) 0.1 = (26.62 + 1.1 X). At the beginning of the fourth year, the contribution will be (26.62 + 2.1 X), and at the end - (26.62 + 2.1 X) + (26,62 + 2,1X) 0.1 = (29.282 + 2.31 X). By condition, you need to find the largest integer x for which the inequality

(29,282 + 2,31x) – 20 – 2x < 17

29,282 + 2,31x – 20 – 2x < 17

0,31x < 17 + 20 – 29,282

0,31x < 7,718

x <	7718
	310

x <	3859
	155

x < 24	139
	155

The largest integer solution to this inequality is the number 24.

Answer: 24.

Task number 18- a task of an increased level of complexity with a detailed answer. This task is intended for competitive selection to universities with increased requirements for the mathematical preparation of applicants. Exercise high level complexity is not a task for applying one solution method, but for a combination of different methods. For the successful completion of task 18, in addition to solid mathematical knowledge, a high level of mathematical culture is also required.

At what a system of inequalities

	x 2 + y 2 ≤ 2ay – a 2 + 1
	y + a ≤ |x| – a

has exactly two solutions?

Solution: This system can be rewritten as

	x 2 + (y– a) 2 ≤ 1
	y ≤ |x| – a

If we draw on the plane the set of solutions to the first inequality, we get the interior of a circle (with a boundary) of radius 1 centered at the point (0, A). The set of solutions of the second inequality is the part of the plane that lies under the graph of the function y = | x| – a, and the latter is the graph of the function
y = | x| , shifted down by A. The solution of this system is the intersection of the solution sets of each of the inequalities.

Consequently, this system will have two solutions only in the case shown in Fig. 1.

The points of contact between the circle and the lines will be the two solutions of the system. Each of the straight lines is inclined to the axes at an angle of 45°. So the triangle PQR- rectangular isosceles. Dot Q has coordinates (0, A), and the point R– coordinates (0, – A). In addition, cuts PR And PQ are equal to the circle radius equal to 1. Hence,

QR= 2a = √2, a =	√2	.
	2

Answer: a =	√2	.
	2

Task number 19- a task of an increased level of complexity with a detailed answer. This task is intended for competitive selection to universities with increased requirements for the mathematical preparation of applicants. A task of a high level of complexity is not a task for applying one solution method, but for a combination of different methods. For the successful completion of task 19, it is necessary to be able to search for a solution, choosing various approaches from among the known ones, modifying the studied methods.

Let sn sum P members of an arithmetic progression ( a p). It is known that S n + 1 = 2n 2 – 21n – 23.

a) Give the formula P th member of this progression.

b) Find the smallest modulo sum S n.

c) Find the smallest P, at which S n will be the square of an integer.

Solution: a) Obviously, a n = S n – S n- 1 . Using this formula, we get:

S n = S (n – 1) + 1 = 2(n – 1) 2 – 21(n – 1) – 23 = 2n 2 – 25n,

S n – 1 = S (n – 2) + 1 = 2(n – 1) 2 – 21(n – 2) – 23 = 2n 2 – 25n+ 27

Means, a n = 2n 2 – 25n – (2n 2 – 29n + 27) = 4n – 27.

B) because S n = 2n 2 – 25n, then consider the function S(x) = | 2x 2 – 25x|. Her graph can be seen in the figure.

It is obvious that the smallest value is reached at the integer points located closest to the zeros of the function. Obviously these are points. X= 1, X= 12 and X= 13. Since, S(1) = |S 1 | = |2 – 25| = 23, S(12) = |S 12 | = |2 144 – 25 12| = 12, S(13) = |S 13 | = |2 169 – 25 13| = 13, then the smallest value is 12.

c) It follows from the previous paragraph that sn positive since n= 13. Since S n = 2n 2 – 25n = n(2n– 25), then the obvious case when this expression is a perfect square is realized when n = 2n- 25, that is, with P= 25.

It remains to check the values ​​​​from 13 to 25:

S 13 = 13 1, S 14 = 14 3, S 15 = 15 5, S 16 = 16 7, S 17 = 17 9, S 18 = 18 11, S 19 = 19 13 S 20 = 20 13, S 21 = 21 17, S 22 = 22 19, S 23 = 23 21, S 24 = 24 23.

It turns out that for smaller values P full square is not achieved.

Answer: A) a n = 4n- 27; b) 12; c) 25.

________________

*Since May 2017, the DROFA-VENTANA joint publishing group has been part of the Russian Textbook Corporation. The corporation also included the Astrel publishing house and the LECTA digital educational platform. Alexander Brychkin, a graduate of the Financial Academy under the Government of the Russian Federation, candidate of economic sciences, head of innovative projects of the DROFA publishing house in the field of digital education (electronic forms of textbooks, Russian Electronic School, LECTA digital educational platform) has been appointed General Director. Prior to joining the DROFA publishing house, he held the position of Vice President for Strategic Development and Investments of the EKSMO-AST publishing holding. Today, the Russian Textbook Publishing Corporation has the largest portfolio of textbooks included in the Federal List - 485 titles (approximately 40%, excluding textbooks for correctional schools). The corporation's publishing houses own the most popular Russian schools sets of textbooks on physics, drawing, biology, chemistry, technology, geography, astronomy - areas of knowledge that are needed to develop the country's production potential. The corporation's portfolio includes textbooks and study guides For elementary school awarded the Presidential Prize in Education. These are textbooks and manuals on subject areas that are necessary for the development of the scientific, technical and industrial potential of Russia.

I will solve the GIA 2017 Grade 9, such search query interested in those who started preparing for the GIA online based on our textbooks, reshebnikov and GDZ. Our site is an archive and collection of all GIA options for last years and contains all the tasks that you will meet at the final certification. Decide OGE options you will have to devote 2 to 3 hours a day to this activity throughout the year. This is necessary to do in order to consolidate the material and accurately control progress at the stage of preparation for the GIA. When we say “I will solve the GIA in mathematics”, then, as a rule, we mean the options that Dmitry Gushchin developed and implemented. It was he who created the site where he collected all the OGE tests in mathematics with answers, where you can conveniently solve OGE options and test your skills in solving equations and problems.

I will pass the GIA online, express preparation for the final certification

I will solve the GIA 2017 online is a proven collection of exercises where you can dwell on a topic in more detail, help a friend and repeat the material covered. The importance of the Russian language in human life is difficult to overestimate. We all communicate, transfer information to each other, read newspapers and tabloid press, where all news articles are written in Russian. That is why we use educational process tests and assignments for preparing for the GIA-2017 in the Russian language. For a deeper perception of information, solution books and workbooks are recommended, which can also be downloaded from our website. I will solve GIA Gushchin Grade 9 is effective way check yourself and the quality of teaching from secondary special educational institutions Russian Federation.

GIA in the Russian language, tests and tasks to prepare for the GIA-2017

Russian language teacher Alevtina Ivanovna Sinichko forced us to solve GIA options online in 2017, when we were young and provocative running through the cool puddles in the courtyard of our school No. 346 in Moscow. She was right, it was necessary to encourage students to study more, solve theorems, prove equations with three unknowns and be proud that you studied at a Moscow school. In those young years, we poorly understood the significance of the GIA in mathematics and hoped that all the pranks would get away with us, but time passed and the day of judgment came when the savior came and said “I will solve the GIA Gushchin” and all the students fell on their faces before him.

Now that decades have passed, we recall with a smile our state final certification and think about how we were afraid, how we hoped for a high GPA, as they dreamed of entering the Lomonosov Moscow State University. Now our outlook on life has changed and we see from the height of the past years that it was enough to say “I will solve the GIA 2017 grade 9 according to Gushchin” and all the doors would be wide open for such a brave applicant. You say "Heresy!", I will answer "Yes!". However, let me remind you that you should learn the answers, otherwise you will not pass the exam.

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The educational portal for preparing for exams according to the Dmitry Gushchin system was originally created as a means of self-control and a way of self-preparation of students for the state final certification. However, the creators of this most useful resource laid in it the mechanisms of interaction with teachers.

In these notes, I share my experience of using resource materials offline - without using computers connected to the Internet. For convenience, the material is presented in the form of a text document and in PDF format.

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To use the preview, create yourself a Google account (account) and log in: https://accounts.google.com

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SAM GIA: I DECIDE OGE and USE educational portal Dmitry Gushchin. User experience

The toolkit offered to the user of the portal RESHU OGE and USE is very wide, and allows you to pass testing both on individual topics and on a complex task. The function of random selection of tasks from the catalog is extremely useful. It is implemented with buttons.homework template And Sample control work .

When the so-called Homework then the advantages of the randomized choice of task are leveled by the fact that each task contains its number. Students on another tab open the catalog and, having found the answer there by this identifier, drive it into the text of the work.

When setting a control work, the number is not affixed, however, in some types of tasks, you can very quickly find the correct answer in the text. Therefore, the resource can currently only be used as a means of preparing students for testing in the classroom.

To obtain objective results of assessing the quality of education, the following applications are possible: testing on the Gushchin website SDAM GIA: I RESOLVE the OGE and the Unified State Examination, testing using other systems or paper media. Experience shows that the only way to conduct objective testing on the described resource is to turn off access to the Internet after the students receive the assignment, and turn it on to send the results. If such technical feasibility If not, then the tasks have to be copied to another testing system. I use MyTestX from A. Bashlakov (http://mytest.klyaksa.net ).

I will move on to the algorithm for using the resource SDAM GIA: I DECIDE the OGE and the USE to create a multivariate verification work. The OGE is a standardized exam, that is, each type of task corresponds to a specific topic, and all you need to do is learn the theory and solve, solve, solve tasks sequentially on each topic, gradually filling your hand to automatism. Some unexpected, or not related to school curriculum There are no questions on the exam. Experience shows that good method training is the solution a large number tasks of the same type, we will create such a test.

Go to the "Teacher" tab

Set the test parameters , from this tab is availableprint version.