Вопрос в картинках...

Question 1

Решите задачу:

$сделать вариант 40$

Question 2

Решите задачу:

$\displaystyle 1. \sqrt{9-x} - 2 = x-5 \\ \\ \sqrt{9-x} = x-3 \\ \\ OD3 \\ x \leq 9 \\ x \geq 3 \\ x \in [3;9] \\ \\ (\sqrt{9-x})^2 = (x-3)^2 \\ \\ 9-x = x^2 - 6x + 9 \\ \\ x^2 - 6x+9+x-9 = 0 \\ \\ x^2 -5x = 0 \\ \\ x(x-5) = 0 \\ \\ x_1 \neq 0 \\ \\ x_2 = 5$

$\displaystyle \sqrt{2x-1} - \sqrt{3x+1} = -1 \\ \\ OD3 \\ \\ \left \{ {{2x-1 \geq 0} \atop {3x+1 \geq 0}} \right. \;\; \left \{ {{x \geq \frac{1}{2} } \atop {x \geq - \frac{1}{3} }} \right. \;\; x \geq - \frac{1}{3} \\ \\ ( \sqrt{2x-1} - \sqrt{3x+1})^2 = (-1)^2 \\ \\ 2x-1 - 2 \sqrt{(2x-1)(3x+1)} + 3x+1 = 1 \\ \\ 2x-1 + 3x +1 -1 = 2 \sqrt{(2x-1)(3x+1)} \\ \\ 5x-1 = 2 \sqrt{(2x-1)(3x+1)} \\ \\ OD3_2 \\ \\ x \geq \frac{1}{5} \\ \\ (5x-1)^2 = 4*(2x-1)(3x+1)$

$\displaystyle (5x-1)^2 = 4*(2x-1)(3x+1) \\ \\ 25x^2 - 10x +1 = 4(6x^2 - 3x + 2x -1) \\ \\ 25x^2 - 10x +1 = 24x^2 - 4x - 4 \\ \\ x^2 - 6x + 5 = 0 \\ \\ D = 36 - 20 = 16 \\ \\ \sqrt{D} = 4 \\ \\ x_1 = \frac{6+4}{2} = 5 \\ \\ x_2 = 1$

$\displaystyle 3. \frac{ \sqrt{x+11} }{2} - \frac{2}{ \sqrt{x+11} } = \frac{3}{2} \\ \\ OD3 \\ x \ \textgreater \ -11 \\ \\ t = \sqrt{x+11} \\ \\ OBP \\ t \geq 0 \\ \\ \frac{t}{2} - \frac{2}{t} = \frac{3}{2} \;\;\;\;\; \bigg|(*2t) \\ \\ t^2 - 4 = 3t \\ \\ t^2 -3t-4 = 0 \\ \\ D = 9+16 = 25 \\ \\ \sqrt{D} = 5 \\ \\ t_1 = \frac{3+5}{2} = 4 = \sqrt{x_1+11} \\ \\ x_1 = 5 \\ \\ t_2 = \frac{3-5}{2} \neq -1 \;\;(! t \geq 0!) \\ \\ OTBET \\ x=5$

Question 3

t - это замена

Question 4

Заменил корень(x+11) на t для удобства