Task number: 43791. Prototype #:
The boat must cross a river with a width of \ (L = 21 \) m and with a current speed \ (u = 0.3 \) m / s so as to moor exactly opposite the place of departure. It can move at different speeds, while the travel time, measured in seconds, is determined by the expression \ (t = \ frac (L) (u) (\ mathop (\ rm ctg) \ nolimits) \ alpha \), where \ ( \ alpha \) - an acute angle that sets the direction of its movement (measured from the coast). What is the minimum angle \ (\ alpha \) (in degrees) you need to swim so that the travel time is no more than 70 s?
Answer:

Task number: 43793. Prototype #:
The boat must cross the river \\ (L = 63 \\) m wide and with a current \\ (u = 1 \\) m / s so that it moors exactly opposite the place of departure. It can move at different speeds, while the travel time, measured in seconds, is determined by the expression \ (t = \ frac (L) (u) (\ mathop (\ rm ctg) \ nolimits) \ alpha \), where \ ( \ alpha \) - an acute angle that sets the direction of its movement (measured from the coast). What is the minimum angle \ (\ alpha \) (in degrees) you need to swim so that the travel time is no more than 63 s?
Answer:

Task number: 43795. Prototype #:
The boat must cross the river \\ (L = 49 \\) m wide and with a current \\ (u = 0.7 \\) m / s so that it moors exactly opposite the place of departure. It can move at different speeds, while the travel time, measured in seconds, is determined by the expression \ (t = \ frac (L) (u) (\ mathop (\ rm ctg) \ nolimits) \ alpha \), where \ ( \ alpha \) - an acute angle that sets the direction of its movement (measured from the coast). What is the minimum angle \ (\ alpha \) (in degrees) you need to swim so that the travel time is no more than 70 s?
Answer:

Task number: 43797. Prototype #:
A skateboarder jumps onto a platform on the rails at a speed of \ (v = 3.2 \) m / s at an acute angle \ (\ alpha \) to the rails. From the push, the platform starts to move at a speed \ (u = \ frac (m) ((m + M)) v \ cos \ alpha \) (m / s), where \ (m = 80 \) kg is the mass of a skateboarder with a skateboard , and \ (M = 240 \) kg is the mass of the platform. What is the maximum angle \ (\ alpha \) (in degrees) you need to jump to accelerate the platform to at least 0.4 m / s?
Answer:

Task number: 43799. Prototype #:
A skateboarder jumps onto a platform on the rails at a speed of \ (v = 2.4 \) m / s at an acute angle \ (\ alpha \) to the rails. From the push, the platform begins to move at a speed \ (u = \ frac (m) ((m + M)) v \ cos \ alpha \) (m / s), where \ (m = 70 \) kg is the mass of a skateboarder with a skateboard , and \ (M = 210 \) kg is the mass of the platform. What is the maximum angle \ (\ alpha \) (in degrees) you need to jump to accelerate the platform to at least 0.3 m / s?
Answer:

Task number: 43801. Prototype #:
A skateboarder jumps onto a platform on the rails at a speed of \ (v = 2.4 \) m / s at an acute angle \ (\ alpha \) to the rails. From the push, the platform starts to move at a speed \ (u = \ frac (m) ((m + M)) v \ cos \ alpha \) (m / s), where \ (m = 80 \) kg is the mass of a skateboarder with a skateboard , and \ (M = 240 \) kg is the mass of the platform. What is the maximum angle \ (\ alpha \) (in degrees) you need to jump to accelerate the platform to at least 0.3 m / s?
Answer:

Task number: 43803. Prototype #:
A skateboarder jumps onto a platform on the rails at a speed of \ (v = 2.4 \) m / s at an acute angle \ (\ alpha \) to the rails. From the push, the platform begins to move at a speed \ (u = \ frac (m) ((m + M)) v \ cos \ alpha \) (m / s), where \ (m = 75 \) kg is the mass of a skateboarder with a skateboard , and \ (M = 225 \) kg is the mass of the platform. What is the maximum angle \ (\ alpha \) (in degrees) you need to jump to accelerate the platform to at least 0.3 m / s?
Answer:

Task number: 43805. Prototype #:
A skateboarder jumps onto a platform on the rails at a speed of \ (v = 2 \) m / s at an acute angle \ (\ alpha \) to the rails. From the push, the platform begins to move at a speed \ (u = \ frac (m) ((m + M)) v \ cos \ alpha \) (m / s), where \ (m = 75 \) kg is the mass of a skateboarder with a skateboard , and \ (M = 225 \) kg is the mass of the platform. What is the maximum angle \ (\ alpha \) (in degrees) you need to jump to accelerate the platform to at least 0.25 m / s?
Answer:

28010. The boat must cross a river with a width of L = 100 m and with a current velocity of u = 0.5 m / s so as to moor exactly opposite the place of departure. It can move at different speeds, while the travel time, measured in seconds, is determined by the expression:

α is an acute angle that sets the direction of its movement (measured from the coast). What is the minimum angle α (in degrees) you need to swim so that the travel time is no more than 200 s.

In order to represent the process of movement, let's build a sketch:

If the boat travels to its destination at an angle of 90 degrees to the shore, it will be swept away by the current and will not arrive at its destination. Therefore, it is necessary to direct it at a certain angle α to the bank towards the river flow. We need to determine the smallest angle α at which t ≤ 200.

The task is reduced to solving the inequality:

Since 0 0< α < 90 0 , то рассматриваем решение неравенства только для первой четверти (то есть, периодичность котангенса не учитываем). Изобразим решение неравенства графически:

Determination of the cotangent: the cotangent of an acute angle in a right-angled triangle is the ratio of the adjacent leg to the opposite one.

Consider the triangle AOB. The cotangent of the AOB angle is equal to one at 45 degrees, and will be less than one when the AO leg is less than the OB leg. This will happen when the AOB angle increases from 45 to 90 degrees, which means 45 0< α < 90 0 .

So, you need to swim at a minimum angle of 45 degrees relative to the coast (choose the smallest angle from the interval).

Answer: 45

Solution.

Material objects that are relevant to the situation described in the problem are: a boat, water in a river, Earth's surface, Earth's gravitational field and air
We will include only a boat in the physical system and will ... consider it as a material point. According to the condition of the problem, the speed of the boat is constant, so its movement can be considered uniform and rectilinear.Since the speeds of the boat and the current are small compared to the speed of light, the classical law of addition of speeds can be used to solve the problem. According to him, the absolute speed of the boat is equal to the geometric sum of its relative and portable speeds. Let us connect the stationary frame of reference with the surface of the Earth, the moving one - with water, therefore the relative velocity is v1, and the portable one - v2.So v = v1 + v2.To switch to the scalar notation, we direct the OX axis along the coast, the OY axis - perpendicular to it, and take the origin at the point O, from which the boat began to move. We will start counting the time at the moment the movement starts.Taking into account the law of addition of the speeds of movement of the boat relative to the coast, r = (v1 + v2) t.
Let's project vector quantities on the OX and OY axes.

At the time when the boat reaches the opposite bank (at t = t1), its coordinates will be: x1 = l, y1 = L, where l is the boat's displacement along the coast, L is the width of the river.

From the second equation we obtain

Solving the Unified State Exam 227 Larin's variant. Detailed solution of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 tasks of the training version of the USE Larina No. 227 (alexlarin.com)

Solving the Unified State Exam 227 Larin's variant. Detailed solution 16,17,18,19 tasks of the training version of the exam Larin No. 227 (alexlarin.com)

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Exercise 1

At school # 1, lessons begin at 8:30, each lesson lasts 45 minutes, all breaks, except one, last 10 minutes, and the break between the second and third lesson is 20 minutes. Now the clock is 13:00. In how many minutes will the next bell from the lesson ring?

To solve the simplest option is to make a schedule for the beginning and end of the lessons:
1)8:30-9:15
1)9:25-10:10
1)10:30-11:15
1)11:25-12:10
1)12:20-13:05
That is, after 5 minutes the bell will ring

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Assignment 2

The figure in bold dots shows the average monthly exchange rate of the RMB from January to August 2014. Months are indicated horizontally, and the yuan price in rubles is indicated vertically. For clarity, bold points are connected with a line. Determine from the figure the difference in the yuan exchange rate in August and July. Give your answer in rubles.

Answer: 0.27

As you can see from the figure, the angle is based on the diameter of the circle, which means that the triangle is rectangular, that is, the answer is $$ 90 ^ (\ circ) $$

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Assignment 4

Anya and Tanya choose one natural number from 1 to 9 independently of each other. Find the probability that the sum of these numbers is divisible by 3. Reduce your answer to hundredths.

Answer: 0.33

Let Anya choose 1, Tanya can choose 9 numbers for this. Similarly, from 2, 3 and so on to 9. That is, there will be a total of 9 * 9 = 81 combinations.
Moreover, in every nine combinations, 3 will be divisible by 3 (since in numbers arranged in a row, every third is divisible by three). That is, 9 * 3 = 27
Then the probability is: $$ P = \ frac (27) (81) = 0, (3) $$
If we round up to hundredths, we get 0.33

Since there is an even root, the radical expression must be greater than or equal to zero. Since there is a variable on the right, and an even root on the left, the function on the right must also be non-negative:
$$ \ left \ (\ begin (matrix) 19 + 6x \ geq 0 \\ x + 4 \ geq 0 \ end (matrix) \ right. \ Leftrightarrow $$$$ \ left \ (\ begin (matrix) x \ geq - \ frac (19) (6) \\ x \ geq -4 \ end (matrix) \ right. $$
Next, let's square both sides:
$$ 19 + 6x = x ^ (2) + 8x + 16 \ Leftrightarrow $$$$ x ^ (2) + 2x-3 = 0 \ Leftrightarrow $$$$ x_ (1) = x_ (2) = - 3 $ $.
Both roots fit in the ODZ, therefore, we choose the smallest one.

If we consider the triangle AOC, then it turns out to be isosceles, since OA = OC - radii. In this case: $$ \ angle AOC = 180 -2 * 37 = 106 ^ (\ circ) $$. But this angle is central, while ∠ABC is inscribed, and then its degree measure is equal to half of the degree measure ∠AOC, that is, 53

The derivative is negative where the function decreases. At all intervals, only one point has an integer abscissa (2; 0)

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Assignment 8

Find the volume of the pyramid shown in the figure. Its base is a polygon, the adjacent sides of which are perpendicular, and one of the side edges is perpendicular to the base plane and is equal to 3.

To solve this problem, the easiest way is to complete the missing part to a regular quadrangular pyramid, find the volume of this pyramid, and subtract the volume of the completed part:
$$ V = \ frac (1) (3) * 6 * 6 * 3 - \ frac (1) (3) * 3 * 3 * 3 = 27 $$

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Assignment 10

The boat must cross a river with a width of L = 100 m so as to moor exactly opposite the place of departure. River flow velocity u = 0.5 m / s. Travel time, measured in seconds, is equal to $$ t = \ frac (L) (u) ctg \ alpha $$, where α is an acute angle between the boat's axis and the coastline. At what minimum angle α to the shore should the boat be directed so that the travel time is no more than 200 s? Give your answer in degrees.

Let's substitute the available data into the equation:
$$ 200 = \ frac (100) (0.5) ctg \ alpha $$
$$ ctg \ alpha = 1 $$
$$ \ alpha = 45 ^ (\ circ) +2 \ pi * n $$, choose the smallest one, this is 45 degrees

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Assignment 11

The cyclist rode the first third of the route at a speed of 12 km / h, the second third at a speed of 16 km / h, and the last third at a speed of 24 km / h. Find the average speed of the cyclist along the entire path. Give your answer in km / h.

Let 3S be the total distance. Then the time in the first section is: $$ t_ (1) = \ frac (S) (12) $$. In the second section: $$ t_ (2) = \ frac (S) (16) $$. In the third segment, the time is: $$ t_ (3) = \ frac (S) (24) $$
The average speed is calculated as the ratio of the total distance traveled to the total time spent: $$ v = \ frac (3S) (\ frac (S) (12) + \ frac (S) (16) + \ frac (S) (24)) = $$$$ \ frac (3S) (\ frac (9S) (48)) = \ frac (3S * 48) (9S) = 16 $$

Find the derivative of this function: $$ y "= \ frac ((2x + 7) * x- (x ^ (2) + 7x + 49)) (x ^ (2)) = $$$$ \ frac (2x ^ (2) + 7x-x ^ (2) -7x-49) (x ^ (2)) = $$$$ \ frac (x ^ (2) -49) (x ^ (2)) = 0 $$ Let's draw a coordinate line, mark the obtained points, and rasterize the signs of the derivative:

As you can see, -7 is the maximum point, therefore, on the interval specified by the condition, at this point there will be the maximum value of the function:

$$ y (-7) = \ frac ((- 7) ^ (2) +7 * (- 7) +49) (- 7) = - 7 $$

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Assignment 13

a) Solve the equation: $$ \ cos 2x +3 \ sqrt (2) \ sin x -3 = 0 $$
b) Indicate the roots of this equation that belong to the segment $$ (\ frac (\ pi) (4); \ pi] $$

Answer: A) $$ (- 1) ^ (k) \ frac (\ pi) (4) + \ pi k, k \ in Z $$ B) $$ \ frac (3 \ pi) (4) $$

A) Apply the double angle cosine formula $$ \ cos 2x = 1-2 \ sin ^ (2) x $$: $$ \ cos 2x + 3 \ sqrt (2) \ sin x-3 = 0 \ Leftrightarrow $$ $ $ 1-2 \ sin ^ (2) x + 3 \ sqrt (2) \ sin x-3 = 0 \ Leftrightarrow $$ $$ 2 \ sin ^ (2) x-3 \ sqrt (2) + 2 = 0 $$

$$ D = (3 \ sqrt (2)) ^ (2) -4 * 4 = 18-16 = 2 $$

Since $$ - 1 \ leq \ sin x \ leq 1 $$, then $$ \ sin x = \ frac (\ sqrt (2)) (2) \ Leftrightarrow $$ $$ x = (- 1) ^ (k ) \ frac (\ pi) (4) + \ pi k, k \ in Z $$

$$ \ left [\ begin (matrix) \ sin x = \ frac (3 \ sqrt (2) + \ sqrt (2)) 4 (= \ sqrt (2)) \\\ sin x = \ frac (3 \ sqrt (2) - \ sqrt (2)) (2) = \ frac (\ sqrt (2)) (2) \ end (matrix) \ right. $$

B) Find the roots of the equation on the interval $$ (\ frac (\ pi) (4); \ pi] $$ using the trigonometric circle: $$ x = \ frac (3 \ pi) (4) $$

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Assignment 14

The base side of a regular triangular prism ABCA 1 B 1 C 1 is equal to $$ 10 \ sqrt (3) $$, and the height of CC 1 is 7.5. A point P is marked on the edge B 1 C 1 so that B 1 P: PC 1 = 1: 3. Points Q and M are the midpoints of sides AB and A 1 C 1, respectively. Plane $$ \ alpha $$ is parallel to line AC and passes through points P and Q.

A) Prove that the line BM is perpendicular to the plane $$ \ alpha $$

B) Find the distance from point M to the plane $$ \ alpha $$

Answer: $$ \ frac (9 \ sqrt (5)) (2) $$

A) 1) $$ a \ cap (ABC) = QT \ left | \ right | AC $$, $$ a \ cap (A_ (1) B_ (1) C_ (1)) = PN \ left | \ right | A_ (1) C_ (1) $$, because $$ a \ left | \ right | AC. a \ cap (BGM) = EF $$, $$ BM \ cap EF = S $$ (E and F are midpoints of PN and QT). BM-oblique, BG-its projection, $$ BG \ perp QT \ Rightarrow $$ for three perpendiculars $$ BM \ perp QT (1) $$

2) $$ \ angle SBF = \ beta $$, $$ \ angle BFS = \ gamma $$, $$ \ angle BSF = \ varphi $$; $$ BG = AB * \ sin 60 = 10 \ sqrt (3) * \ frac (\ sqrt (3)) (2) = 15 $$; $$ tg \ beta = \ frac (MG) (BG) = \ frac (7,5) (15) = \ frac (1) (2) $$; $$ ctg \ gamma = \ frac (\ frac (1) (2) BF) (BB_ (1)) = $$$$ \ frac (1) (4) * \ frac (15) (7,5) = $$$$ \ frac (1) (2) = tg \ beta \ Rightarrow $$ $$ \ beta + \ gamma = 90 $$, then $$ \ varphi = 90 $$, $$ BM \ perp EF (2 ) $$. From (1) and (2) $$ \ Rightarrow $$ $$ BM \ perp \ alpha $$

B) 1) from item a) $$ BM \ perp \ alpha \ Rightarrow $$ $$ p (Ma) = MS $$

2) $$ \ Delta ESM \ sim \ Delta FSB $$ in two corners $$ \ Rightarrow $$ $$ \ frac (MS) (BS) = \ frac (ME) (BF) = \ frac (3) (2 ) $$, then $$ MS = \ frac (3) (5) BM $$; $$ BM = \ sqrt (BG ^ (2) + MG ^ (2)) = \ sqrt (225+ \ frac (225) (4)) = \ frac (15 \ sqrt (5)) (2) $$ , $$ MS = \ frac (3) (5) * \ frac (15 \ sqrt (5)) (2) = \ frac (9 \ sqrt (5)) (2) $$

The range of permissible values ​​of inequality is set by the system:

$$ \ left \ (\ begin (matrix) 10-x ^ (2)> 0 \\ 10-x ^ (2) \ neq 1 \\\ frac (16) (5) xx ^ (2)> 0 \ end (matrix) \ right. \ Leftrightarrow $$ $$ \ left \ (\ begin (matrix) - \ sqrt (10)

Solution: $$ \ log_ (10-x ^ (2)) (\ frac (16) (5) x-x ^ (2))<1\Leftrightarrow$$ $$\left[\begin{matrix}\left\{\begin{matrix}10-x^{2}>1 \\\ frac (16) (5) x-x ^ (2)<10-x^{2}\end{matrix}\right.\\\left\{\begin{matrix}0<10-x^{2}<1\\\frac{16}{5}x-x^{2}>10-x ^ (2) \ end (matrix) \ right. \ End (matrix) \ right. \ Leftrightarrow $$ $$ \ left [\ begin (matrix) \ left \ (\ begin (matrix) -3 3 \\ x<-3\end{matrix}\right.\\\frac{16}{5}x>10 \ end (matrix) \ right. \ End (matrix) \ right. \ Leftrightarrow $$$$ \ left [\ begin (matrix) \ left \ (\ begin (matrix) -3 3 \\ x<-3\end{matrix}\right.\\x>\ frac (25) (8) \ end (matrix) \ right. \ end (matrix) \ right. \ Leftrightarrow $$ $$ \ left [\ begin (matrix) -3

Taking into account the range of admissible values ​​of the inequality, we obtain $$ x \ in (0; 3) \ cup (\ frac (25) (8); \ sqrt (10)) $$

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Task 16

A circle of radius $$ 2 \ sqrt (5) $$ is drawn through the vertices A and B of the triangle ABC, cutting off the segment $$ 4 \ sqrt (5) $$ from the straight line BC and tangent to the straight line AC at point A. From point B, a perpendicular is drawn to the straight line BC before the intersection with the line AC at point F.

A) Prove AF = BF

B) Find the area of ​​the triangle ABC if BF = 2.

Answer: $$ \ frac (5 \ sqrt (5)) (3) $$

By the condition $$ OA = R = 2 \ sqrt (5); BK = 4 \ sqrt (5) $$. Rice. 2 can only be used to prove item a) since by the condition $$ BF = 2 $$, $$ OA = 2 \ sqrt (5) $$, i.e. Bf

a) AC-tangent $$ \ Rightarrow $$ $$ OA \ perp AC, BF \ perp OB, OB = R \ Rightarrow $$ BF-tangent and by the tangent property $$ AF = BF $$

b) 1) Let $$ FC = x, BC = y $$, then $$ AC = x + 2 $$, $$ OC = y + 2 \ sqrt (5) $$

2) $$ \ Delta FBC \ sim OAC $$ in two corners $$ \ Rightarrow $$ $$ \ left \ (\ begin (matrix) \ frac (BF) (OA) = \ frac (BC) (AC) \ \\ frac (BF) (OA) = \ frac (FC) (OC) \ end (matrix) \ right. \ Leftrightarrow $$$$ \ left \ (\ begin (matrix) \ frac (2) (2 \ sqrt (5)) = \ frac (y) (x + 2) \\\ frac (2) (2 \ sqrt (5)) = \ frac (x) (y + 2 \ sqrt (5)) \ end (matrix ) \ right. \ Leftrightarrow $$$$ \ left \ (\ begin (matrix) y = \ frac (x + 2) (\ sqrt (5)) \\ y = \ sqrt (5) (x-2) \ end (matrix) \ right. \ Leftrightarrow $$$$ \ left \ (\ begin (matrix) x = 3 \\ y = \ sqrt (5) \ end (matrix) \ right. $$

$$ FC = 3, BC = \ sqrt (5), AC = 5 $$, $$ \ frac (S _ (\ Delta ABC)) (s _ (\ Delta BFC)) = \ frac (AC) (FC) = \ frac (5) (3) $$;

$$ S _ (\ Delta BFC) = \ frac (1) (2) BC * BF = \ sqrt (5) $$ then, $$ S _ (\ Delta ABC) = \ frac (5) (3) $$, $$ S _ (\ Delta BFC) = \ frac (5 \ sqrt (5)) (3) $$

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Task 17

Vasya dreams of his own apartment, which costs 3 million rubles. Vasya can buy it on credit, while the bank is ready to issue this amount immediately, and Vasya will have to repay the loan in equal monthly installments for 20 years, while he will have to pay an amount 180% higher than the original. Instead, Vasya can rent an apartment for some time (the rental price is 15 thousand rubles per month), postponing every month for the purchase of an apartment the amount that will remain from his possible payment to the bank (according to the first scheme) after paying the rent for the rented apartment ... For how many years, in this case, will Vasya be able to save up for an apartment, if we assume that its cost will not change?

Answer: 12.5

The apartment costs 3 (million rubles) = 3000 (thousand rubles), the loan is taken for 20 (years) = 240 (months). We will solve the problem by actions:

1) 3000 * 2.8 = 8400 (thousand rubles) - the total amount of payments to the bank;

2) 8400: 240 = 35 (thousand rubles) - monthly payment to the bank;

3) 35-15 = 20 (thousand rubles) - the amount that Vasya can save every month after paying the rent;

4) 3000: 20 = 150 (months) = 12.5 (years) - Vasya will need to save up for an apartment.

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Task 18

Find all values ​​of the parameter a for each of which the system $$ \ left \ (\ begin (matrix) 1- \ sqrt (| x-1 |) = \ sqrt (7 | y |) \\ 49y ^ (2) + x ^ (2) + 4a = 2x-1 \ end (matrix) \ right. $$ has exactly four different solutions.

Answer: $$ - \ frac (1) (4); - \ frac (1) (32) $$

Let's rewrite the system as $$ \ left \ (\ begin (matrix) \ sqrt (\ left | x-1 \ right |) + \ sqrt (7 \ left | y \ right |) = 1 \\\ left | x- 1 \ right | ^ (2) + (7 \ left | y \ right |) ^ (2) = - 4a \ end (matrix) \ right. $$

Let $$ \ sqrt (\ left | x-1 \ right |) = m \ geq 0 $$; $$ \ sqrt (7 \ left | y \ right |) = n \ geq 0 $$

Then the system will take the form: $$ \ left \ (\ begin (matrix) m + n = 1 \\ m ^ (4) + n ^ (4) = - 4a \ end (matrix) \ right. (*) $$ . If the pair of numbers $$ (m_ (0); n_ (0)) $$ is a solution to the system (*), then the pair $$ (n_ (0); m_ (0)) $$ is also its solution:

1) Let $$ m_ (0) \ neq n_ (0), m_ (0), n_ (0)> 0 $$. Then $$ \ left [\ begin (matrix) \ left \ (\ begin (matrix) \ left | x-1 \ right | = m_ (0) ^ (2) \\ 7 \ left | y \ right | = n_ (0) ^ (2) \ end (matrix) \ right. \\\ left \ (\ begin (matrix) \ left | x-1 \ right | = n_ (0) ^ (2) \\ 7 \ left | y \ right | = m_ (0) ^ (2) \ end (matrix) \ right. \ end (matrix) \ right. (**) $$ Each system in an aggregate has four solutions, then a given system has 8 different solutions that do not satisfy the condition of the problem.

2) Let one of the values ​​$$ m_ (0) $$ or $$ n_ (0) $$ be equal to zero, then pairs (0; 1) and (1; 0) -solutions of the system (*), -4a = 1 , whence $$ a = - \ frac (1) (4) $$. In this case, the set (**) will take the form:

$$ \ left [\ begin (matrix) \ left \ (\ begin (matrix) \ left | x-1 \ right | = 0 \\ 7 \ left | y \ right | = 1 \ end (matrix) \ right. \\\ left \ (\ begin (matrix) \ left | x-1 \ right | = 1 \\ 7 \ left | y \ right | = 0 \ end (matrix) \ right. \ end (matrix) \ right. $$, whence we get 4 solutions of this system: $$ (1; \ frac (1) (7)) $$, $$ (1; - \ frac (1) (7)) $$, $$ (2; 0) $$, $$ (0; 0) $$

3) Let $$ m_ (1) = n_ (0) $$, then $$ \ left \ (\ begin (matrix) m_ (0) + m_ (0) = 1 \\ m_ (0) ^ (4) + m_ (0) ^ (4) = - 4a \ end (matrix) \ right. $$., whence

$$ m_ (0) = \ frac (1) (2) $$, $$ a = - \ frac (1) (32) $$ and system (*) has one solution $$ (\ frac (1) ( 2); \ frac (1) (2)) $$. In this case, the set (**) will take the form:

$$ \ left \ (\ begin (matrix) \ left | x-1 \ right | = \ frac (1) (4) \\ 7 \ left | y \ right | = \ frac (1) (4) \ end (matrix) \ right. $$, whence we get 4 solutions of this system: $$ (1 \ frac (1) (4); \ frac (1) (28)) $$, $$ (1 \ frac (1) (4); - \ frac (1) (28)) $$, $$ (\ frac (3) (4); \ frac (1) (28)) $$, $$ (\ frac (3) ( 4); - \ frac (1) (28)) $$.

Let us prove that for $$ a = - \ frac (1) (4) $$ and $$ a = - \ frac (1) (32) $$ this system has no other solutions except the found solutions.

1. For $$ a = - \ frac (1) (4) $$ system (*) has the form: $$ \ left \ (\ begin (matrix) m + n = 1 \\ m ^ (4) + n ^ (4) = 1 \ end (matrix) \ right. $$. If $$ m \ neq 0 $$, $$ n \ neq 0 $$, then $$ m, n \ in (0; 1) $ $ and $$ \ left \ (\ begin (matrix) m ^ (4)

Then $$ m ^ (4) + n ^ (4)

2. For $$ a = - \ frac (1) (32) $$ system (*) has the form: $$ \ left \ (\ begin (matrix) m + n = 1 \\ m ^ (4) + n ^ (4) = \ frac (1) (8) \ end (matrix) \ right. $$. Let $$ \ left \ (\ begin (matrix) m = \ frac (1) (2) + t \\ n = \ frac (1) (2) -t \ end (matrix) \ right. $$, then $$ \ left \ (\ begin (matrix) m ^ (4) = (\ frac (1) (2) + t) ^ (2) = \ frac (1) (16) + 4 * \ frac (1) (8) t + 6 * \ frac (1) (4) t ^ (2) + 4 * \ frac ( 1) (2) t ^ (3) + t ^ (4) \\ n ^ (4) = (\ frac (1) (2) -t) ^ (4) = \ frac (1) (16) - 4 * \ frac (1) (8) t + 6 * \ frac (1) (4) t ^ (2) -4 * \ frac (1) (2) t ^ (3) + t ^ (4) \ end (matrix) \ right. $$. And $$ m ^ (4) + n ^ (4) = \ frac (1) (8) + 3t ^ (2) + 2t ^ (4) $$. We have: $$ \ frac (1) (8) + 3t ^ (2) + 2t ^ (2) = \ frac (1) (8) $$, whence $$ t = 0 $$, $$ m = n = \ frac (1) (2) \ Rightarrow $$ there are no other solutions and $$ a = - \ frac (1) (32) $$ satisfies the condition.

Answer: 1,3, (5); no; 8

Let us denote the differences from the problem statement by $$ s_ (1) $$ and $$ s_ (2) $$, the nth term of the progression - by $$ x_ (n) $$, the sum of its first n terms - by $$ S_ (n) $$. As you know, the square of the sum of any number of terms is equal to the sum of squares and various doubled products of terms. Therefore: $$ s_ (1) = 2 (x_ (1) x_ (2) + ... + x_ (n-1) x_ (n)) $$, $$ s_ (2) = 2 (x_ (1 ) x_ (2) + .. + x_ (n) x_ (n + 1)) $$. $$ s_ (2) $$ includes all terms from $$ s_ (1) $$ and doubled products $$ x_ (n + 1) $$ for all terms of the progression from $$ x_ (1) $$ to $$ x_ (n) $$. Hence, $$ s_ (2) -s_ (1) = 2x_ (n + 1) (x_ (1) + .. x_ (n)) = 2x_ (n + 1) S_ (n) (1) $$

A) Answer: 1,3, (5). If $$ s_ (2) -s_ (1) = 40, x_ (n + 1) S_ (n) = 20 $$. The last equality holds, for example, for the progression 1,3, (5).

B) Answer: I could not. Under the conditions of the problem, the smallest value in (1) for n = 13 is $$ 2 * 13 (0 + 1 + .. + 12) = 2028> 1768 $$

C) Answer: 8.From formula (1) we get: $$ s_ (2) -s_ (1) = 2x_ (n + 1) \ frac ((x_ (1) + x_ (n)) n) (2) = x_ (n + 1) (x_ (1) + x_ (n)) n = 1768 $$. Therefore, $$ 1768 = 2 ^ (3) * 13 * 17 $$ is divisible by n. From point B) $$ n<13$$ ; наибольший из таких делителей равен 8 . Проверим это значение. Если $$n=8$$, $$x_{9}(x_{1}+x_{8})=13*17$$. Возможны следующие два варианта

1. $$ x_ (9) = 17 \ Rightarrow $$ $$ x_ (8) \ leq 13 \ Rightarrow $$ progression difference $$ d \ geq 4 \ Rightarrow $$ $$ x_ (1) = x_ (9) -8d \ leq 17-32<0$$

2. $$ x_ (9) = 13 \ Rightarrow $$ for $$ d \ geq 2 $$ we will have: $$ x_ (1) = x_ (9) -8d \ leq 13-16<0$$. Значит, $$d=1$$. Конечная прогрессия 5,6,7,8,9,10,11,12 удовлетворяет условию задачи.