For the first time in my life, I took the CT in Russian on June 15, 2014. I have had a higher education for a long time: when I entered, there were regular exams, not CT. At the university, at the Faculty of Journalism, I also studied the rules of Russian grammar and stylistics. But since during my work in the media proofreaders corrected my mistakes, my knowledge was thoroughly erased from my memory.

In recent years, when I began to fulfill copywriting orders, and then teach copywriting, the forgotten began to be lacking. Congenital literacy, i.e. good visual memory, nature did not reward me, so when checking my own and other people's texts, I have to recall the rules from the textbook. A copywriter is his own proofreader, and students sometimes put commas anywhere and wait for well-grounded comments. Therefore, I decided to repeat the subject, using the CT as a stimulus. And in addition to improving the skill of competent writing, it was important for me to get the results of another experiment, on the independent organization of the educational process.

At the first attempts to fill in the tests of the past years, I had from 5 to 10 errors: an indicator that is usual for a good student who did not prepare specifically for testing. But, unlike the middle school student, I was sure: it is quite possible to prepare for the exam with high quality in a short time. The experiment was a success: I passed 100 points.

What is a "short term" for preparation for CT in Russian?

The guide I used to prepare for CT

I decided that I would take the CT, and for the first time opened the test preparation guide on April 26, 2014. The last days of April and the whole of May I did not more than 4 hours a week. It took a little more than a month to read, without stopping at the exercises, "A manual for preparation for compulsory centralized testing" O. Gorbatsevich, T. Ratko and T. Bodarenko and to perform several tests over the previous year. At the end of May, I bought an additional collection of tests for 2008-2012 and from June 1 to June 14 I studied for 3 hours a day, dealing in detail with the most difficult topics and solving tests for different years.
Thus, the total number of hours devoted exclusively to preparation for DH is about 60, in addition, travel time in transport was sometimes used. All this time, I continued to work and do my household chores.

What was easier for me than for applicants?

If you compare me with a high school graduate, then I won in that I was not burdened with final exams. During the same period, the schoolchildren had only 2-3 days to distract themselves from the examination subjects and concentrate all their attention on the CT. Compared to the average “adult” applicant, I won in the fact that I had a free work schedule and equal sharing of household chores with my spouse: this means that I could choose the most productive time to prepare for the tests. In addition, I took one, not three tests.
Others will have to adjust to their circumstances. This means, most likely, that the repetition will take more calendar days. On the other hand, if you start preparing in advance, then your knowledge will be more solid. Then the final repetition just before testing may take only a few hours.

What was more difficult for me than for applicants?

They say that with age, the tenacity of memory weakens, and I partly believe in this. In addition, I did not have Russian language lessons at school and preparation for a school exam in the same subject. The latter circumstance has both pros and cons. Because school assignments on the subject also allow you to consolidate the topics brought up on the Central Television in your memory.

I didn't have counseling teachers, electives, tutors, and applicants' courses. Nobody organized my educational process. Although the latter circumstance did not bother me, because it was the technologies of self-organization that I tested on myself.

How can I benefit from my experience preparing for VG?

These are the things that seem common to me.

1. Maximum program and optimum program
   For a test, the results of which are assessed in points, it is easy to define your tasks in the same units. I had two slats. I wanted to take the maximum: precisely in order to talk about this result in confirmation of the possibilities of rational self-education. But the goal of “100 points and not less” is counterproductive. There is an element of randomness in any business, so, I decided, it depends on me to pass the test by at least 90 points. In this case, I will be able to share my experience. If it is less, it means that testing will remain my personal experiment. The maximum program, perhaps, can be the same for everyone: if someone passes the CT at 100, then you can. The problem with the optimum. Applicants do not know it, as a pragmatically valuable value, because you cannot predict a competition for a specific university in advance. I don't think it's worth exercising in predictions. The optimum is the score that you yourself will consider an achievement: it just needs to be higher than your "normal level".
2. Motivation
   I was motivated by the desire to write this text, and then invite those who wish to take a course on self-organization in studies. After all, I allowed myself to do this only if I successfully passed the test. But besides this goal, connected with the repetition of the school curriculum only indirectly, there was also a simple everyday benefit from preparing for the CT. I write texts to order all the time, as well as correct mistakes of novice copywriters. And constantly checking the textbooks as you work is a waste of time. As I assume, applicants want to do very strongly, and this desire can stimulate efforts to prepare for the CT. But it’s impossible to predict in advance what grade will be enough for admission, and this is discouraging. Maybe, if you more clearly link preparation for tests with the acquisition of practical literacy, the desire to understand the rules will be higher? True, I don't know what everyday tasks to associate a test in mathematics or chemistry with, but I suppose that if you find out in advance how a particular science is used in a particular future profession, then you can find meaning in drawing up chemical equations. Among the factors of motivation, scientists single out and procedural and meaningful, emphasizing their importance. In other words, to work with enthusiasm, one must be inspired not only by future results, but also by the process of work itself. I had no problems with this: it is interesting for me to understand how the language works. If there is no such curiosity, and there is still enough time to prepare, you can try to awaken a keen interest in linguistics in yourself. It didn't work out - well, increase external stimuli, come up with additional rewards for your studies.
3. Solid surface repetition and in-depth study of difficult sections
   As I said before, at first I was just leisurely reading the DH preparation guide. And in the process of reading, I noted which sections seem confusing and indistinct to me. While performing tests in parallel, I noticed questions that I could not answer confidently. The first candidates for a second repetition and study of additional sources were topics that were difficult to answer both in answering the questions and in the course of reading the manual. In second place is something that was not easy to understand when reading. But the section "phonetics" was enough to re-read (and write down the number of letters and sounds) once. It is somehow self-sufficient, and schoolchildren make the least mistakes in it.
4. Repetition sequence
   Russian (and Belarusian for those who regularly use it) are perhaps the most practical subjects in the school curriculum. After all, we all write some texts, and writing them competently is a simple rule of decency. Therefore, the value of the sections "spelling" and "punctuation" is seen by all conscious applicants. But the sections "morphology", "word formation" and "syntax" at first glance are far from the tasks of everyday life. But after all, its spelling directly depends on the composition of the word and on what part of speech it belongs to, and the arrangement of punctuation marks depends on the structure of the sentence. Therefore, it is worth starting with "impractical" sections and dealing with them as carefully as possible. And spelling will grow on the basis of this knowledge with almost no effort.
5. Sources for repetition
   The manual that I have used is considered by the tutors to be one of the best, but it cannot be the only one. To fully understand complex topics, alternative sources are needed. There are many of them on the Internet, it is important to be able to choose reliable ones. And if you don’t know how, learn.
6. Mnemonics
   On the VKontakte network there is a popular group called Memories of the Russian Language. One or two such didactic works have been firmly engraved in the memory since school: we will refer to the second conjugation without a doubt ... By the way, do you remember that in this rhyme - not all verbs are not in-letter, referring to the second conjugation? It does not include verbs with a stressed ending in the third person plural, for example, "to burn." This I mean that it is not worth replacing the understanding of the issue with mnemonics. Although there are things that really need to be remembered, and here the mnemonics will help us. For the most confusing rules, I drew “rule cards” (according to the technique described in the book by EE Vasilieva and V.Yu. Vasiliev “Super memory, or how to remember to remember”), to memorize lists (for example, the categories of pronouns by meaning ) used the Loki technique (binding a word to a familiar place). I wrote down vocabulary words with double consonants on several sheets of paper and hung them on the door, difficult to pronounce, recorded them on a dictaphone and listened to them in the transport. To test my memorization, I recruited my daughter several times to give me a vocabulary dictation. It seems that's all.
7. Self-test
   When filling out the "practice" tests, I, of course, checked the filling result. Moreover, I did not check the answers with the answers during the first check. And almost every time she herself found 1-2 mistakes "due to inattention".
   For some reason, I think that most people have such mistakes. If I had more time to prepare, I would definitely include in the program separate exercises for training attention. But a simple self-test is quite effective. By the way, in my real exam test I found a missing mark in my self-test.
   I was very surprised, watching my colleagues on the delivery of the CT: almost everyone who left before me did not reread their work at all. Why? After all, correct corrections here are guaranteed to increase the score!
8. Audience strategies
   They ask to come for testing in half an hour, but the envelopes with tests are opened exactly at 11. The meaning of half an hour of sitting at a desk is not very clear. There is enough time for the test, even under the stress of the exam, and most of the test takers left long before the final call. Phones really need to be handed over to the wardrobe, all handbags, pieces of paper and even glasses cases are collected and generally kept vigil. You can't write off.
   But you can write comments to the questions. I wrote directly to two of my fourth version of the test. There was a grammatical error in the wording of one of the questions, the second (according to the analysis of the text) was, in my opinion, incorrectly formulated. But whether my last remark was taken into account or my guesses in the test form simply coincided with what the authors had in mind, remained a mystery to me. One thing is obvious: it is safe to write comments on tests, this does not affect the score. Most likely, these comments do not correlate with the authorship of the form either, because they asked to draw up their views “in a shorter way, so that they could be included in the protocol” and did not require a signature.
9. After testing
   I didn’t want to “take the test and forget” and after a few days I realized the need for regular repetition. Successfully selected cells in the test form are not yet ideal literacy and not even expert knowledge of a school course.
   The rules that are used in almost every text are well remembered: punctuation for homogeneous members of a sentence, endings of verbs and nouns, spelling of frequently used vocabulary words. For the rest, you have to periodically refer to dictionaries and reference books. I got into the habit of clarifying questionable spelling points after completing work on each article, installed an app for spaced repetitions on my smartphone - so I don't have to flip through dictionaries too often.
   Differences between conjunctions and allied words, types of circumstances and categories of pronouns are gradually forgotten. But my "rule cards" are in a safe place, so that the material in memory can now be renewed not in a month and a half, but in a day. And this is good news: my experience may be useful to others more than once.

How self-organization andlearningskills?

In the senior grades of American schools and in the first years of universities and colleges, many students take courses in learning skills, educational skills. We don't have such practice, so I studied this subject on my own. These are general planning and time management skills, as well as the ability to extract the main thing from the text (by the way, it helps with text analysis tasks), take notes, work in a group and individually, memorize different types of information. Before experimenting on myself, I had already conducted several motivational groups for self-organization in everyday life, and one for managing personal projects in any areas: some of the skills of self-organization relate to study in the same way as to regular menu planning.

So, here are the training skills that were used in preparation.

1. Time planning, scheduling
   On the last day, all the preparations were gone, which is good. It is possible to remember a large amount of information per day with a developed RAM, but it is unlikely to learn how to apply the rules.
2. Sequence planning
   The sections in the textbooks are not in the most convenient order. But if you have already made Gantt charts a couple of times for sensible renovation of a room or filling a site, then it's easier to figure out the sequence of topics you personally need to repeat.
3. Critical thinking
   This is also one of the training skills that must be applied in time for the choice of information sources. On the Internet there are both reliable resources for self-study, and pages that are outwardly quite convincing, but reflect the necessary data only partially. If you do not distinguish the former from the latter, you can miss important points.
4. Correct self-study procedures
   On any specialized resource for preparing for VG, you will find many tips, the meaning of which is obvious. Write down the main thing on paper, use different senses, read more ... The devil is in the details: how exactly to write this "main thing" so that it remains not only on paper, what is worth saying, but what is better to practice once again on tests, which texts are useful "Read more", in what tasks this reading can help. For these procedures, Americans take learning skills courses.
5. Periodic maintenance processes
   Standard procedures that should be followed in order not to lose, in this case, the knowledge gained. These procedures are also necessary in the course of study, especially extended over a long period. But now they still need to be adapted to the situation of real use of knowledge.

Can I learn from my experience in preparing for DH?

I think you can. And I personally find it more interesting that it is useful not so much for filling out test forms as for improving practical literacy. After all, this is the skill that will be needed all your life. And together with it, in a natural way, you can also learn general educational skills, the very learning skills with which no exam is terrible.

This idea formed the basis for a course of practical literacy, which I already propose today to individually take any high school student and adult. The general route is similar: to determine what you don't know, to choose from the whole body of knowledge exactly those that you need - and to study consistently using the tools of effective self-education. Meet the program and come!
,

How to pass the CT in Russian for 100 points? Personal experience

How to get a high score on CT and USE?

See below how to successfully prepare for the exam from scratch, repeat and deepen your knowledge, solve all the tasks of the CT, RT and USE of previous years, problems of all types, as well as get video solutions to score more than 70 points on the CT and USE

To successfully pass the CT and USE

You can access:

1. To the video solution of ALL types of problems in chemistry

2.To tests and complete video solutions of CG of all years (starting from the very first CG in 2004), demo tests of different years from the RIKZ website, as well as different stages of RT (starting from 2013) + trial versions of CG 2020

3. To ALL materials on the site, including all tests and assignments with video explanations, courses on problem solving, trial versions of the CT and USE 2020 ( full access)

As a result of obtaining full access, you:

• You will know the most effective ways to complete the tasks of the CT and USE, which 70% of schoolchildren do not know;
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• Solve all tasks of the CT and USE;
• Get all the literature you need;
• You will be able to successfully pass the CT and Unified State Exam, enter the university you dreamed of;
• Make a total breakthrough in the development of your skills and knowledge;
• You solve more than 3000 author's tests and tasks, as well as more than 300 unique problems that are guaranteed to help you cope with the exam and CT in chemistry.

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Frequently Asked Questions about video preparation:

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Yes, of course they will. The courses are designed in such a way that you will learn how to solve problems from the simplest to the most difficult. That is, even if you absolutely do not know how to solve problems. The main thing is to solve and watch the video explanations of all the objectives of the course.

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Option 1

Part B

Task B1.For painting the walls with a total area of \u200b\u200b175 m 2, it is planned to purchase paint. The volume and cost of paint cans are shown in the table.

What is the minimum amount (in rubles) to spend on the purchase of the required amount of paint if its consumption is 0.2 l / m 2?

Decision.

Since 1 m 2 leaves 0.2 liters of paint, then 175 m 2 will require a volume of paint equal to 175 0.2 \u003d 35 liters.

Thus, the challenge is to find a minimum purchase price of 35 liters or more of paint.

Let's determine the cost of 1 liter of paint in each of the cans.

The price of a liter in a 2.5 liter can is 75,000: 2.5 \u003d 30,000 rubles, and the price of a liter in a 10 liter can is 270,000: 10 \u003d 2,700 rubles.

Since paint is cheaper in large cans, it is advisable to collect 35 liters of paint using only large cans. However, you cannot get exactly 35 liters with large cans, since each of the cans has a volume of 10 liters. There are two options here:

1. We buy 4 cans of paint, 10 liters each. As a result, we have 40 liters of paint, which exceeds the 35 liters we need. The price of paint in this case: 270,000 4 \u003d 1,080,000 rubles.

2. We buy 3 cans of paint, 10 liters each and 2 cans of paint, 2.5 liters each. As a result, we have exactly 35 liters of paint. The price of paint in this case: 3 · 270,000 + 2 · 75,000 \u003d .960,000 rubles.

Since the second option is cheaper than the first, the minimum amount required to buy the required amount of paint is 960,000 rubles.

Answer:960 000.

Do you have any questions or comments on solving the problem? Ask the author, Anton Lebedev.

Task B2.Find the sum of the roots (root, ec is he the only one) equations

Decision.

First, note that squaring both sides of the equation is not a good idea in this task, since as a result we get an equation of degree 4, which in general cannot be solved

In such situations, workarounds should be sought.

To begin with, we define the ODZ equation:

The resulting equation is equivalent to the system:

Comment.The first inequality of the system is necessary in order to avoid the appearance of unnecessary roots: if we simply square both sides, then the roots of the equation will also be added to the roots of the equation.

So, we solve the equation from the written system:

Obviously, only the second of the found roots satisfies the inequality from the system.

Thus, the original equation has only one root, which is 9.

Answer:9.

Task B3.A circle is inscribed in an isosceles trapezoid, the area of \u200b\u200bwhich is equal. The sum of the two angles of a trapezoid is 60 °. Find the perimeter of the trapezoid.

Decision.

Let ABCD Is a given trapezoid.

Since the trapezoid is isosceles, the angles at the base of the trapezoid are equal:

.

By condition, the sum of the two angles of the trapezoid is 60 °. Obviously, we are talking about two sharp corners, since 60 °< 90° , which means that in our notation we are talking just about the cornersBAD and CDA ... Since they are equal, and their sum is 60 °, then each of them is equal to 30 °.

As you know, not every trapezoid (and not every isosceles trapezoid) can be inscribed with a circle, which means that the fact that a circle is inscribed in our trapezoid gives us some additional information. A circle can be inscribed only in a trapezoid in which the sum of the bases is equal to the sum of the sides. In our case, it should be:

Since the trapezoid is isosceles, then AB \u003d CD... Let us denote the lateral sides by x.

Then we get

where MN - the middle line of the trapezoid.

The height of the VK trapezoid is also expressed through x... To do this, consider a right-angled triangleABK.

.

Then the sum of the sides is 2 x \u003d17, and the perimeter of the trapezoid is 34 (the sum of the bases equals the sum of the sides).

Answer:34.

Task B4.Let (x, y) - solution of the system of equations

Find the meaning of the expression 5y - x.

Decision.

We transform the second equation of the system:

Taking into account the first equation, we get:

We calculate the value of the expression:

Answer:23.

Task B5.Find the meaning of the expression

Decision.

Comment.The most frequent problems of applicants when solving such examples is the inability to get rid of the irrationality in the denominator by multiplying by the conjugate and the ignorance of the fact that the order of calculation of consecutive roots does not matter (for example,).

Answer:-22.

Task B6.Find the sum of the roots of the equation.

Decision.

Before starting the solution, we say the magic phrase: "The product is equal to zero if at least one of the factors is equal to zero." After that, the equation miraculously decomposes into a set:

The first equation in the population has a single root x = 81.

Let's transform the second equation:

The further solution is carried out by changing the variable:

We get

(roots were found using Vieta's converse theorem).

The negative root does not suit us, so we get

This means that the original equation has two roots: 1 and 81.

Their sum is 82.

Answer:82.

Task B7.Find the area of \u200b\u200bthe lateral surface of a regular triangular pyramid if the length of the bisector of its base is equal and the plane angle at the apex is.

Decision.

Let SABC - regular triangular pyramid.

Triangle ABC - the base of the pyramid, and this triangle is regular.

The bisector is also the height of the triangle ABC, therefore

The lateral surface area of \u200b\u200ba regular pyramid is S \u003d SK· p,

where

- base semi-perimeter;

Apothem.

Then

S = 125 \u003d 60.

Answer:60.

Problem B8.Find the sum of the smallest and largest integer solutions to the inequality

Decision.

Considering that the logarithm is an increasing function if its base is greater than 1 and decreasing if its base is less than 1, as well as the fact that the sub-logarithm expression must be positive, we get:

The smallest integer solution is -5, and the largest is 65. Their sum is 60.

Answer:60.

Task B9.Find (in degrees) the sum of the roots of the equation 10sin5 xCos5 x + 5sin10 xCo18 x \u003d 0 in the interval (110 °; 170 °).

Decision.

Using the double argument formula, we transform the first term on the left side:

Since from all found roots it is necessary to choose those of them that lie on the interval (110 °; 170 °), then

We write out the corresponding roots:

126 °; 144 °; 162 °

130 °; 150 °.

The sum of the solutions found is 712.

Answer:712.

Task B10.Find the product of the smallest and largest integer solutions to the inequality

Decision.

We transform the original inequality:

The resulting inequality can be solved, for example, by the method of intervals. To do this, we first find the roots of the corresponding equation:

The found roots are plotted on the number axis. These roots split the expression (| x + 5| - 4)(|x - 3 | - 1) for intervals of constancy. Determine the sign of the written expression at each of the intervals by substituting any point from the specified interval into the expression. For example, to determine the sign of the expression on the rightmost interval, take the point x \u003d 5 and we get that the value of the expression at this point is positive, which means that the expression will be positive over the entire interval.

Now we can write down the solution to the inequality (the corresponding area is shaded in the figure):

.

Smallest integer from this range: x min \u003d -8, and the largest integer x max \u003d 3. The product of these numbers is -8 · 3 \u003d -24. This number should be written in the answer.

Answer:-24.

Task B11.Point A moves along the perimeter of the triangleKMP. Points K1 , M1, P1 lie on the medians of the triangleKMP and divide them in the ratio of 11: 3, counting from the top. Along the perimeter of the triangle K1 M1 P1 point B moves at a speed five times greater than the speed of point A. How many times point B goes around the perimeter of the triangle K1 M1 P1 in the time during which point A will go around the triangle twiceKMP.

Decision.

Let's make a drawing for the task. О is the point of intersection of the medians of the original triangle.

Intuitively, triangles KMP and K1 M1 P1 should be similar. However, intuition only suggests a way to solve the problem, so the similarity of these triangles still needs to be proved.

To prove the similarity, consider the triangles KOM and K1 OM1 .

MM ’ Is the median of the triangleKMP , therefore, since the medians of the triangle are divided in a ratio of 2 to 1, counting from the vertex.

It follows from the problem statement that since point M1 divides the median MM 'in a ratio of 11 to 3, counting from the top.

Then

Attitude

.

Similarly, one can show that

Besides, as vertical.

So triangles KOM and K1 OM1 are similar in two sides and the angle between them with a coefficient of similarity.

Then

Similarly

.

This means that the triangles KMP and K1 M1 P1 are similar with the coefficient of similarity and the perimeter of the triangle KMPtimes the perimeter of the triangle K1 M1 P1 .

Since point B moves at a speed 5 times the speed of point A along a triangle, the perimeter of which is one times less than the perimeter of the triangleKMP, then during one revolution of point A, point B makes revolutions, and during two revolutions of point A, point B will make 56 revolutions.

Answer:56.

Task B12.Volume of a rectangular parallelepiped ABCDA1 B1 C1 D1 equals 1728. Point P lies on the side edge CC1 so that CP:PC1 \u003d 2: 1. Through point P, topD and the middle of the lateral rib AA1, a cutting plane is drawn, which divides the rectangles into two parts. Find the volume of the smaller part.

Decision.

Draw a parallelepiped in the drawing and build the described sectionPDKEF. K - mid rib AA1 .

Draw in the drawing the lines along which the section plane intersects the planes of the three faces of the parallelepiped. Points where the section plane intersects straight lines BA, BC and BB1 denote by Z, Q, S.

Body SZBQ - pyramid, at the base of which lies a right-angled triangleZBQ ... This pyramid includes the volume of the lower part of the parallelepiped and the volumes of the three pyramids SEB1 F, QPCD, ZKAD.

To find the volume of the lower part of the parallelepiped, we find the volumes of the indicated pyramids.

For convenience of calculations, we denote the sides of the parallelepiped by x, y and z, then the volume of the parallelepiped V = xyz = 1728.

Besides,

.

The task is to express the sizes of these four pyramids in terms of x, y and z.

Triangles FC1 P and DAK are similar in two angles (all sides of these triangles are parallel in pairs).

Then

.

Triangles PCD and KA1 E are also similar, therefore

.

From the similarity of triangles SB1 F and PC1 F follows:

.

Pyramid volume SEB1 F is equal to:

Pyramid QPCD like a pyramid SEB1 F with the similarity coefficient:

.

Then the volume of the pyramid QPCDis equal to:

Similar to pyramid ZKAD like a pyramid SEB1 F with the similarity coefficient

Then the volume of the pyramid ZKADis equal to:

Finally, the pyramid SZBQ like a pyramid SEB1 F with the similarity coefficient

.

Then the volume of the pyramid SZBQis equal to:

The volume of the lower part of the parallelepiped:

Then the volume of the upper part:

Since we need less volume, the correct answer is 724.

Answer:724.

Every teenager sooner or later begins to think about admission and testing. However, the question "How to successfully pass the CT in Russian?" remains open. We present to your attention proven ways to prepare for CT and the best tips from students and teachers to help you pass Russian as "excellent".

CT in Russian

Start your path to success before the rest!

Arriving in grade 10, many think that they have a whole year to rest. But this is not the case. Giving yourself a head start, you have no idea how you will complicate the next year, and how much you will regret the lost time after. After all, for grade 10 you do not have to go through the entire program non-stop. When there is a free minute - spend it usefully. Review the rules, do tests, play video tutorials. So, in the first year, you can repeat a larger part of the program, leaving a small part for the graduation class. And while your classmates are just starting to think about tutors, you are already one step ahead! You are already moving towards your goal!

Rehearsals for the most important event in school life

Have you ever taken a rehearsal test (RT)? I didn't go because you're afraid to find out your results ?! In vain! The sooner you start your rehearsals, the better. The special atmosphere of the RT will help to better transfer the excitement to the CT. If you start going to RT from the beginning of the 10th grade, then there is a possibility that you will not get confused on the exam itself, and the excitement can generally come to naught.

Ways to pass the test successfully

If you take the question on a large scale, you can single out all two ways to prepare for testing:

• with a tutor;
• without a tutor.

And basically many people choose the first option, but is it justified? For self-preparation you need, first of all, patience and perseverance. If you are a happy owner (s) of these two qualities, then this option is for you!

Anastasia Karatkevich, student of Minsk State Linguistic University:

« I prepared for testing myself, because I think it is very real. You just need self-control and that's it».

Preparation materials for DH

The most important assistant in preparation - collection of tests for past years... When preparing, you need to do as many tests as possible. However, you should not solve them at random or, as everyone likes to do at school, in your head. It is necessary to prescribe each task for better assimilation, so that in the end there are no problems at the main event of the whole school life.

Anna Tikhonovich, student of the Institute of Journalism, BSU:

"Towards central heating in Russian, I wouldn’t have prepared much, because I never had any problems with the language. I bought a collection of assignments for 5 or 6 years and solved everything from start to finish. Then I analyzed incomprehensible tasks».

Galina Borodina, teacher of Russian language and literature:

"You can find many good Internet portals where you can get advice and work on thematic simulators."

Few tests from past years and want to prepare more? Check it out on the internet! On the vastness of the world wide web, you can find tests that will immediately show whether you answered correctly and explain everything with the rules.

What to do if you are an ardent fan (-ka) social networks? The answer is simple - join

Option 1

Part B

Problem B1... If in a regular quadrangular pyramid the height is 4, and the area of \u200b\u200bthe diagonal section is 12, then its volume is….

Decision.

Let SABCD- regular quadrangular pyramid with apex Sand the basis ABCD... Quadrangle ABCDis a square, since the pyramid is correct. Triangle BSD- the diagonal section of the pyramid. SO - the height of the pyramid.

The area of \u200b\u200ba diagonal section can be determined using the formula for the area of \u200b\u200ba triangle:

By the condition of the problem, a.

The side of the square is one times smaller than its diagonal (prove it yourself), therefore

Then the area of \u200b\u200bthe base of the pyramid:

The volume of the pyramid is calculated by the formula:

Answer: 24.

Task B2. Find the number of all integer solutions to the inequality

Decision.

We transform the original inequality:

Note that the reduction by x was not specially carried out, since first it is necessary to take into account the ODZ inequality:.

Then the initial inequality on the ODZ is equivalent to the following:

We solve this inequality by the method of intervals, taking into account the ODD.

Then the solution to the inequality has the form:.

We write out the whole solutions included in this area: - 14 solutions in total.

Answer: 14

Task B3.Points A (1; 2), B (5; 6) and C (8; 6) - the tops of the trapezoid ABCD (AD || BC)... Find the sum of the coordinates of a point D, if .

Decision.

Let's designate points A, B and FROM on the coordinate plane. Points B and C lie on a straight line parallel to the abscissa axis. As AD || BCthen a straight line ADmust be parallel to the abscissa axis, which means that the ordinate of the point D is equal to the ordinate of the point. A Let the abscissa of the point D equals x, that is, the coordinates of the point D (x; 2).

For determining x use the fact that. Based on the formula for the distance between two points of the coordinate plane:

From the two obtained values, we choose x \u003d 9since point Dobviously lies to the right of the point A.

Thus, the coordinates of the point D (9; 2)... Their sum is 11. This is the number and must be written in the answer.

Answer: 11

Problem B4... Find the perimeter of a regular hexagon whose smaller diagonal is equal to.

Decision.

Let ABCDEF- considered regular hexagon with sides equal to a . Is the smaller diagonal of this hexagon. Consider a triangle ABC... This triangle has (like the sides of a hexagon) and an angle (prove it yourself).

Apply the cosine theorem to this triangle:

Since, we get:

Then the perimeter of the hexagon

Answer: 60

Task B5.Find the product of the roots of the equation .

Decision.

Let's transform the original equation:

Product of roots:

Answer: -3

Task B6.The area of \u200b\u200bthe rectangle ABCD is 20. Points M, N, P, Q are the midpoints of its sides. Find the area of \u200b\u200ba quadrilateral enclosed between lines AN, BP, CQ, DM

1. Since MB || PD and MB \u003d PDas half of the opposite sides of a rectangle, then MBPD - parallelogram, which means BP || MD... It can be proved similarly that NA || CQ... Hence, in the quadrangle RSTU the opposite sides are parallel and this quadrilateral is a parallelogram.

2. To find the area of \u200b\u200bthe parallelogram, apply the formula:

Where h - the height of the parallelogram lowered to the side RU.

Thus, the task is reduced to finding h and RU.

3. Find the height h... Note that the height h parallelogram RST is also the height of the parallelogram MBPD, so

Parallelogram area MBPD find as the difference of the area of \u200b\u200bthe rectangle ABCD and two right-angled triangles: AMD and CPB:

Since is the area of \u200b\u200bthe rectangle ABCDthen

4. Find RU.

Corner sides ADM intersect by parallel lines AN and QC... As AQ \u003d QD, then by Thales' theorem UD \u003d RU.

Similarly BS \u003d STand since ST \u003d RU (opposite sides of the parallelogram, then BS \u003d RU.

In a triangle BAS line segment MR parallel to the base BS and divides the side AB in half. Means MR - the middle line of the triangle BAS and .

5. Calculate the area of \u200b\u200bthe quadrangle RSTU:

Answer: 4

Task B7. Solve the equation and find the sum of its roots.

Decision.

Let's transform the equation:

Here, we have factorized the quadratic trinomials on the left and right sides of the equation (check that this factorization is actually valid).

Let's denote ODZ:

.

Let's introduce a replacement:

We get the equation:

The first equation of the set of roots has no roots, since its discriminant is negative. The second equation has two roots. According to Vieta's theorem, the sum of these roots is 9.

REMARKS TO THE PROBLEM.

At first glance it seems that it is simply impossible to see all the necessary transformations and replacements on your own. In fact, everything is not so scary. The first thing that catches your eye when solving an equation is two square trinomials on the right and left sides of the expression. At a subconscious level, any schoolchild should already have it laid down: if you see a square trinomial - consider its discriminant. Usually, if the root of the discriminant is easy to calculate, then the solution of the problem is tied to the factorization of the square trinomial, which is done by the first action.

The second step of the solution is parenthesis expansion and replacement. It is difficult to see this transformation straight away if you do not know that it exists, so solve more examples: experience comes with quantity.

Answer: 9.

Problem B8. Find the value of the expression if .

Since, then ... Then .

.

Answer: -6

Task B9. Find the sum of integer values xbelonging to the domain of the function

Decision.

Let us write down the system of inequalities that define the domain of the specified function. For this, we use the properties of the logarithm:

Let's solve the last inequality of the system.

We represent the solution to the inequality on the number line using the method of intervals. On the same line, we put the solutions of the remaining inequalities of the system.

We write down the scope of the function:

Let us write out the whole solutions that are included in the domain of definition:

The sum of these decisions is -6.

Answer: -6

Task B10.A right-angled triangle with legs equal to 6 and rotates around the axis containing its hypotenuse. Find the value of the expression where V Is the volume of the rotation figure.

Decision.

Let ABCIs a given right-angled triangle, and ASIs the hypotenuse of this triangle. Let us depict in the drawing the body obtained by rotating this triangle around the axis containing its hypotenuse.

The resulting body can be divided into two cones, at the base of which lies the same circle centered at the point O... The radius of this circle OB coincides with the height of a right-angled triangle.

Let h Is the height of the triangle. Then the area of \u200b\u200bthe bases of the cones:

Let's write an expression for the volume of the body:

Thus, the task was reduced to finding the hypotenuse ASand heights h triangle ABC.

Hypotenuse ASfind using the Pythagorean theorem:

To find the height h consider right triangles ABCand BOC... These triangles have an angle C common, therefore they are similar in two angles (right and angle C).

From the similarity of triangles it follows:

In this way .

Then the volume of the body:

In response, you must write down the value: