Question

Решите одно уравнения и два неравества.

Answer 1

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

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

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

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

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