Question

3.2

4.2
Пожалуйста решитеее

Answer 1

Ответ:

$1)\ \ (x^2+4x+4)(1-x)\leq 0\\\\(x+2)^2(1-x)\leq 0\ \ ,\ \ \ +++[-2\ ]+++[\ 1\ ]---\\\\\underline {\ x\in \{-2\}\cup [\ 1\ ;+\infty \, )\ }\\\\\\2)\ \ \dfrac{x-2}{x+1}+\dfrac{1}{x-1}=\dfrac{6}{x^2-1}\ \ ,\ \ \ \ ODZ:\ x\ne \pm 1\ ,\\\\\\\dfrac{(x-2)(x-1)+(x+1)-6}{(x-1)(x+1)}=0\ \ ,\ \ \ \dfrac{x^2-3x+2+x+1-6}{(x-1)(x+1)}=0\ \ ,\\\\\\\dfrac{x^2-2x-3}{(x-1)(x+1)}=0\ \ ,\ \ \dfrac{(x+1)(x-3)}{(x-1)(x+1)}=0\ \ ,\ \ \ \dfrac{x-3}{x-1}=0\ \ ,\\\\\\x-3=0\ \ ,\ \ \ \underline {\ x=1\ }$

$3)\ \ y=\sqrt{x^2-3x-10}-\dfrac{5}{x^2-9}\\\\\\OOF:\ \left\{\begin{array}{l}x^2-3x-10\geq 0\\x^2-9>0\end{array}\right\ \ \left\{\begin{array}{l}(x+2)(x-5)\geq 0\\(x-3)(x+3)>0\end{array}\right\\\\\\\left\{\begin{array}{l}x\in (-\infty ;-2\ ]\cup [\ 5;+\infty )\\x\in (-\infty ;-3)\cup (3;+\infty )\end{array}\right\ \ \ \Rightarrow \ \ \ x\in (-\infty ;-3\, )\cup [\ 5\, ;+\infty \, )$