Question

Решите всю часть А и В срочно

Answer 1

А1

$\frac{3x-x^2}{2} + \frac{2x^2-x}{6} = x\\\\9x - 3x^2 + 2x^2 - x = 6x\\\\-x^2 + 2x = 0\\\\x(-x + 2) = 0\\\\x_1 = 0\\\\x_2 = 2$

Правильный ответ №1.

А2

$\frac{x^2}{2-x} = \frac{3x}{2 -x}\\\\ODZ: x\neq2\\\\x^2 = 3x\\\\x^2 - 3x = 0\\\\x(x - 3) = 0\\\\x_1 = 0\\\\x_2 = 3\\\\x_1 + x_2 = 3$

Правильный ответ № 3.

А3

$\frac{2x^2+3x}{3-x} = \frac{x-x^2}{x-3}\\\\ \frac{2x^2+3x}{3-x} = \frac{x^2-x}{3-x}\\\\ODZ: x\neq 3\\\\2x^2 + 3x = x^2 - x\\\\2x^2 - x^2 + 3x + x = 0\\\\x^2 + 4x = 0\\\\x(x+4) = 0\\\\x_1 = 0\\\\x_2 = -4\\\\x_1 + x_2 = 0 + (-4) = 0 - 4 = -4$

Правильный ответ № 3.

A4

$\frac{2y-5}{y+5} = \frac{3y+21}{2y - 1}\\\\ODZ: y \neq -5; y \neq 0,5\\\\(2y-5)(2y-1) = 3(y+7)(y + 5)\\\\4y^2 - 2y - 10y + 5 = 3y^2 + 15y + 21y + 105\\\\4y^2 - 3y^2 - 12y - 36y - 100 = 0\\\\y^2 - 48y - 100 = 0\\\\D = 2304 + 400 = 2704\\\\\sqrt D = 52\\\\y_1 = \frac{48 - 52}{2} = \frac{-4}{2} = -2\\\\y_2 = \frac{48+52}{2} = \frac{100}{2} = 50\\\\\frac{y_1+y_2}{2}= \frac{50-2}{2} = \frac{48}{2} = 24$

Правильный ответ №4

А5

$\frac{3x^2 - 5x - 2}{2-x} = 0\\\\ODZ: x\neq 2\\\\3x^2 - 5x - 2 = 0\\\\D = 25 + 24 = 49\\\\x_1 = \frac{5-7}{6} = \frac{-2}{6} = -\frac{1}{3}\\\\x_2 = \frac{5+7}{6} = \frac{12}{6} = 2$

ОДЗ запрещает нам использовать корень $x =2$

Правильный ответ №3

В1

$\frac{y-14}{y^3-8} = \frac{5}{y^2 + 2y + 4} - \frac{1}{y-2}\\\\ODZ: y^3 - 8\neq0\\\\y^3\neq8\\\\y \neq 2 \\\\ y - 14 = 5(y - 2) - y^2 - 2y - 4\\\\ y - 14 = 5y - 10 - y^2 - 2y - 4\\\\ y^2 - 2y + 5y + y - 10 - 4 + 14 = 0\\\\y^2 + 4y = 0\\\\y(y + 4) = 0\\\\y_1 = 0\\\\y_2 = -4\\\\\frac{y_1+y_2}{2} = \frac{0-4}{2} = \frac{-4}{2} = -2$

B2

$5x = 6 + \frac{4}{x-1}\\\\5x(x - 1) = 6(x - 1) + 4\\\\5x^2 - 5x = 6x - 6 + 4\\\\5x^2 - 11x + 2 = 0\\\\D = 121 - 40 = 81\\\\x_1 = \frac{11-9}{10} = \frac{2}{10} = \frac{1}{5} = 0,2\\\\x_2= \frac{11+9}{10} = \frac{20}{10} = 2\\\\y_1 = 5\cdot 0,2 = 1\\\\y_2 = 5\cdot 2 = 10$

Графики функций $y = 5x$ и $y = 6 + \frac{4}{x-1}$ пересекаются в точках с координатами $(0,2; 1)$ и $(2; 10)$.

Удачи!