Question

решите неравенства. Помогите пожалуйста ​

Answer 1

$1)\ \ \sqrt{\dfrac{x-3}{3-2x}}>-1\\\\Tak\ kak\ \ \sqrt{\dfrac{x-3}{3-2x}}\geq 0\ ,\ \ to\ \ \ \dfrac{x-3}{3-2x}\geq 0\ .\\\\x-3=0\ \ \to \ \ x_1=3\\\\3-2x=0\ \ \to \ \ x_2=1,5\\\\znaki\ drobi:\ \ \ ---(1,5)+++(3)---\\\\Otvet:\ \ x\in (\ 1,5\ ;\ 3\ )\ .$

$2)\ \ \sqrt{x^2-2x-3}<1\ \ \ \Leftrightarrow \ \ \ \left\{\begin{array}{l}x^2-2x-3\geq 0\\x^2-2x-3<1^2\end{array}\right\\\\\\\left\{\begin{array}{l}x\in (-\infty ;-1\, ]\cup [\, 3;+\infty )\\x^2-2x-4<0\end{array}\right\ \ \left\{\begin{array}{l}x\in (-\infty ;-1\, ]\cup [\, 3;+\infty )\\x\in (\, 1-\sqrt5\, ;\, 1+\sqrt5)\end{array}\right\ \ \ \Rightarrow \\\\\\Otvet:\ \ x\in (\ 1-\sqrt5\ ;\ -1\ ]\cup [\ 3\, ;\ 1+\sqrt5\ )\ . \\\\\\\star \ \ x^2-2x-3=0\ \ \to \ \ x_1=-1\ ,\ x_2=3\ \ (teorema\ Vieta)$

$(x+1)(x-3)\geq 0\ \ \ \ znaki:\ +++[-1\, ]---[\, 3\, ]+++\\\\x\in (-\infty ;-1\, ]\cup [\ 3\, ;+\infty )\ \ \star \\\\\star \ \ x^2-2x-4=0\ \ ,\ \ D/4=1+4=5\ \ ,\ \ x_{1,2}=1\pm \sqrt5\\\\x^2-2x-4<0\\\\znaki\ \ (x^2-2x-4):\ +++(1-\sqrt5)---(1+\sqrt5)+++\\\\x\in (\ 1-\sqrt5\ ;\ 1+\sqrt5\ )\ \ \star$

$3)\ \ 0<x+\sqrt{2-x}\ \ \ \Rightarrow \ \ \sqrt{2-x}>-x\ \ \ \Leftrightarrow \\\\\\\left\{\begin{array}{l}-x\geq 0\\2-x>(-x)^2\end{array}\right\ \ \ ili\ \ \ \left\{\begin{array}{l}-x<0\\2-x\geq 0\end{array}\right\\\\\\\left\{\begin{array}{l}x\leq 0\\x^2+x-2<0\end{array}\right\ \ \ ili\ \ \ \left\{\begin{array}{l}x>0\\x\leq 2\end{array}\right$

$\left\{\begin{array}{l}x\leq 0\\x\in (-2\, ;\, 1\, )\end{array}\right\ \ \ ili\ \ \ \ \ x\in (\ 0\ ;\ 2\ ]\\\\\\x\in (-2\ ;\ 0\ ]\ \ \ \ ili\ \ \ \ x\in (\ 0\ ;\ 2\ ]\\\\Otvet:\ x\in (-2\ ;\ 2\ ]\ .\\\\\star \ \ x^2+x-2<0\ \ ,\ \ x_1=-2\ ,\ x_2=1\\\\znaki:\ \ +++(-2)---(1)+++\ \ \ x\in (-2\, ;\ 1\ )\ \ \star$