Question

Допомжіть, помогитееееее, пжж​

Answer 1

Ответ:

$1)\ \ (2-x)\sqrt{3+2x-x^2}\geq 0\ \ \Rightarrow \ \ \ \left\{\begin{array}{l}2-x\geq 0\\3+2x-x^2\geq 0\end{array}\right\ \ \left\{\begin{array}{l}x\leq 2\\x^2-2x-3\leq 0\end{array}\right\\\\\\\left\{\begin{array}{l}x\leq 2\\(x+1)(x-3)\leq 0\end{array}\right\ \ \left\{\begin{array}{l}x\in (-\infty ;2\ ]\\x\in [-1;\ 3\ ]\end{array}\right\ \ \ \ \Rightarrow \ \ \underline {\ x\in [-1\ ;\ 2\ ]\cup \{3\}\ }$

$2)\ \ |2x^2-12x+13|\geq 3\ \ \Leftrightarrow \ \ \left[\begin{array}{l}2x^2-12x+13\geq 3\\2x^2-12x+13\leq -3\end{array}\right\ \ \left[\begin{array}{l}2x^2-12x+10\geq 0\\2x^2-12x+16\leq 0\end{array}\right$

$\left[\begin{array}{l}x^2-6x+5\geq 0\\x^2-6x+8\leq 0\end{array}\right\ \ \left[\begin{array}{l}(x-1)(1-5)\geq 0\\(x-2)(x-4)\leq 0\end{array}\right\ \ \left[\begin{array}{l}x\in (-\infty ;\ 1\ ]\cup [\ 5\ ;+\infty )\\x\in [\ 2\ ;\ 4\ ]\end{array}\right\\\\\\\underline {\ x\in x\in (-\infty ;\ 1\ ]\cup [\ 2\ ;\ 4\ ]\cup [\ 5\ ;+\infty ]\ }$