Question

Решить пожалуйста интервалов

Answer 1

$xne 2; ,; xne -1$

$frac{x+4}{x-2} textless frac{2}{x+1};$

$frac{x+4}{x-2}-frac{2}{x+1} textless 0\\frac{(x+4)(x+1)-2(x-2)}{(x-2)(x+1)} textless 0\\frac{x^2+5x+4-2x+4}{(x-2)(x+1)} textless 0\\frac{x^2+3x+8}{(x-2)(x+1)} textless 0\\x^2+3x+8 textgreater 0; ,; ; t.k.; ; D=9-4cdot 8 =-23 textless 0; ; Rightarrow \\(x-2)(x+1) textless 0\\+++(-1)---(2)+++\\xin 9-1,2)$

$2); ; frac{6}{2-4x}-frac{5x}{1-2x} leq 0\\frac{2cdot 3}{2(1-2x)}-frac{5x}{1-2x} leq 0\\\frac{-(5x-3)}{-(2x-1)} leq 0; ,; ; xne frac{1}{2}\\+++(frac{1}{2})---(frac{3}{5})+++\\xin (frac{1}{2},frac{3}{5}, ]$

$3); ; frac{2}{x^3}-frac{3}{x^2} textless frac{5}{x}; ,; ; xne 0\\frac{2-3x-5x^2}{x^3} textless 0\\frac{5x^2+3x-2}{x^3} textgreater 0\\5x^2+3x-2=0; ,; ; D=49; ,; x_1=frac{-3-7}{10}=-1; ,; x_2=frac{4}{10}=frac{2}{5}\\---(-1)+++(0)---(frac{2}{5})+++\\xin (-1,0)cup (frac{2}{5},+infty )$