Question

Помогите пожалуйста решить неравенства

Answer 1

$1)\; \; \dfrac{x^2\, (6-x)}{x^2-4x+4}\leq 0\; \; \; \to \; \; \; \dfrac{x^2(x-6)}{(x-2)^2}\geq 0\\\\\\x_1=0\; ,\; x_2=6\; ,\; x_3=2\\\\znaki:\; \; \; ---[\, 0\, ]---(\, 2\, )---[\, 6\, ]+++\\\\\underline {\; x\in \{0\}\cup [\, 6;+\infty )\; }$

$2)\; \; \left\{\begin{array}{l}7-2x\geq x-2\\x^2-2x-63>0\end{array}\right\; \; \left\{\begin{array}{l}9\geq 3x\\(x-9)(x+7)>0\end{array}\right\; \; \left\{\begin{array}{l}x\leq 3\\x\in (-\infty ;-7)\cup (9;+\infty )\end{array}\right\\\\\\\underline {\; x\in (-\infty ;-7\, )\; }\; \; \; -\; otvet\\\\\star \; \; x^2-2x-63>0\; \; ,\; \; D/4=1+63=64\; ,\; x_1=-7\; ,\; x_2=9\; \; \; \to\\\\x^2-2x-63=(x-9)(x+7)>0\\\\znaki:\; \; +++(-7)---(9)+++$