Question

Решить неравенство с модулем ​

Answer 1

$\displaystyle \tt \frac{x(x-3)^2}{|x+4|}\leq 0\:\:\:\:\: | \: x\ne-4\\\\\displaystyle \tt \frac{x(x^2-6x+9)}{|x+4|}\leq 0\\ \\\displaystyle \tt \frac{x^3-6x^2+9x}{|x+4|}\leq 0\\\\\\\displaystyle \tt \frac{x^3-6x^2+9x}{x+4}\leq 0, \: \: x+4\geq 0\\\\\displaystyle \tt \frac{x^3-6x^2+9x}{-(x+4)}\leq 0, \: \: x+4\leq 0\\\\\\\displaystyle \tt x\in(-4;\:0]\cup \lbrace 3\rbrace, \: \: x\geq -4\\\displaystyle \tt x\in(-\infty;\:-4)\cup[0;\:+\infty), \: \: x<-4$

$\displaystyle \tt x\in(-4;\:0]\cup\lbrace 3 \rbrace\\\displaystyle \tt x\in(-\infty;\:-4)\\\\\\\displaystyle \tt x\in(-\infty;\:-4)\cup(-4;0]\cup\lbrace 3 \rbrace , \: \: x\ne-4\\\\\\\displaystyle \tt x\in(-\infty;-4)\cup(-4;\:0]\cup\lbrace 3 \rbrace$