Решите одно уравнения и два неравества.

Question 1

Question 2

Решите задачу:

$1)\quad \sqrt{\frac{x+1}{x}} -3 \sqrt{ \frac{x}{x+1} } -2=0\; ,\; \; ODZ:\; x\in (-\infty ,-1)\cup (0,+\infty )\\\\t=\sqrt{\frac{x+1}{x}} \geq 0\; ,\; \; \; t-\frac{3}{t}-2=0\; ,\; \; \frac{t^2-2t-3}{t}=0\; \; \to \\\\t^2-2t-3=0\; \; \to \; \; t_1=-1,\; t_2=3\\\\ \sqrt{\frac{x+1}{x} }=-1\; \; net\; \; reshenij\\\\\sqrt{\frac{x+1}{x}}=3\; ,\; \; \frac{x+1}{x}=9\; ,\; \; x+1=9x\; ,\; \; 8x=1\; ,\; \boxed {x=\frac{1}{8}}$

$2)\quad \sqrt{\frac{3x-2}{x+1} } -10\cdot \sqrt{\frac{x+1}{3x-2} } +3 \geq 0\; ,\; \; ODZ:\; x\in (-\infty ,-1)\cup (\frac{2}{3},+\infty )\\\\t=\sqrt{\frac{3x-2}{x+1}} \geq 0\; ,\; \; t-\frac{10}{t}+3 \geq 0\; ,\; \frac{t^2+3t-10}{t} \geq 0\\\\ \frac{(t+5)(t-2)}{t} \geq 0\quad ---[-5\, ]+++(0)---[\, 2\, ]+++\\\\t\in [-5,0)\cup [\, 2,+\infty )\\\\No\; \; t \geq 0\; \; \to \; \; t\in [\, 2,+\infty ))\; \; \to \; \; \sqrt{\frac{3x-2}{x+1}} \geq 2\; \; \Leftrightarrow$

$\left \{ {{\frac{3x-2}{x+1} \geq 0} \atop {\frac{3x-2}{x+1} \geq 4}} \right. \; \left \{ {{x\in (-\infty -1)\cup [\, 2,+\infty )+ODZ} \atop {\frac{3x-2-4x-4}{x+1} \geq 0}} \right. \; \left \{ {{x\in (-\infty ,-1)\cup (\frac{2}{3},+\infty )} \atop {\frac{-x-6}{x+1} \geq 0}} \right. \to \\\\ \left \{ {{x\in (-\infty ,-1)\cup (\frac{2}{3},+\infty )} \atop {x\in [-6,-1)}} \right. \; \; \Rightarrow \; \; \boxed {x\in [-6,-1)}$

$3)\quad \frac{3x-2}{2x-3}- \sqrt{ \frac{3x-2}{2x-3} } -6 \geq 0 \; ,\; ODZ:\; x\in (-\infty ,\frac{2}{3}\, ]\cup (\frac{3}{2},+\infty )\\\\t=\sqrt{\frac{3x-2}{2x-3}} \geq 0\; ,\; \; \; t^2-t-6 \geq 0\ ,\; (t+2)(t-3) \geq 0\\\\+++[-2]---[3]+++\\\\t\in (-\infty ,-2\, ]\cup [\, 3,+\infty )\in ODZ\\\\No\; \; t \geq 0\; \; \to \; \; t\in [\, 3,+\infty )\; \; \to \; \; \sqrt{\frac{3x-2}{2x-3}} \geq 3\\\\\frac{3x-2}{2x-3} \geq 9\; ,\; \; \frac{3x-2-9(2x-3)}{2x-3} \geq 0\; ,\; \; \frac{-15x+25}{2x-3} \geq 0\; ,$

$\frac{-5(3x-5)}{2x-3} \geq 0\; ,\; \; \frac{5(3x-5)}{2x-3} \leq 0\\\\+++(\frac{3}{2})---[\frac{5}{3}\, ]+++\\\\\boxed {x\in \left (\frac{3}{2},\frac{5}{3}\, \right ]}$