Решить пожалуйста интервалов. Created by batya07. algebra-ru

Решить пожалуйста интервалов...

batya07 @batya07

July 2021 1 20 Report

Решить пожалуйста интервалов

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NNNLLL54

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$x\ne 2\; ,\; x\ne -1$

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

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

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

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

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