Question

Решите неравенства. Помогите пожалуйста, очень надо​

Answer 1

$1)\ \ x\cdot \sqrt{x^2-x-2}\geq 0\ \ ,\ \ \ \ \ \ \ \star \ \ A\sqrt{B}\geq 0\ \ \Leftrightarrow \ \ \left[\begin{array}{l}B=0\\\left\{\begin{array}{l}A\geq 0\\B\geq 0\end{array}\right\end{array}\right\ \star \\\\\\\left[\begin{array}{l}x^2-x-2=0\\\left\{\begin{array}{l}x\geq 0\\x^2-x-2\geq 0\end{array}\right\end{array}\right\ \ \ \left[\begin{array}{l}x_1=-1\ ,\ x_2=2\\\left\{\begin{array}{l}x\geq 0\\x\in (-\infty ;-1\, ]\cup [\, 2\, ;+\infty )\end{array}\right\end{array}\right$

$\left[\begin{array}{l}x_1=-1\ ,\ x_2=2\\x\in [\, 2\, ;+\infty )\end{array}\right\ \ \ \Rightarrow \ \ \ \ Otvet:\ \ x\in \{-1\}\cup [\, 2\, ;+\infty )\ .$

$2)\ \ \sqrt{3x-4}>\sqrt{4-x}\ \ ,\ \ \ \ \star \ \ \sqrt{A}>\sqrt{B}\ \ \Leftrightarrow \ \ \left\{\begin{array}{l}A>0\\B\geq 0\end{array}\right\ \star \\\\\\\left\{\begin{array}{l}3x-2>0\\4-x\geq 0\end{array}\right\ \ \left\{\begin{array}{l}3x>2\\4\geq x\end{array}\right\ \ \left\{\begin{array}{l}x>\dfrac{2}{3}\\x\leq 4\end{array}\right\ \ \ \Rightarrow \ \ \ \ Otvet:\ \ x\in \Big(\ \dfrac{2}{3}\ ;\ 4\ \Big]$

$3)\ \ \sqrt{x^2}+x<1\ \ \ \Rightarrow \ \ \ |x|+x<1\\\\a)\ \ x\geq 0:\ |x|=x\ \ ,\ \ x+x<1\ \ ,\ \ 2x<1\ \ ,\ \ x<0,5\\\\b)\ \ x<0:\ |x|=-x\ \ ,\ \ -x+x<1\ \ ,\ \ 0<1\ \ \underline {verno}\ \ pri\ \ x\in (-\infty ;+\infty )\\\\Otvet:\ \ x\in (-\infty \, ;\ 0,5\ )\ .$