Question

Срочноооооооооо ребятааааа помогитеее с полным решением

Answer 1

Ответ:   23 - Г  ,  24 - Б  .

$23)\ \ \ cos^2x=5+5sinx\ \ ,\ \ x\in \Big[-\dfrac{\pi}{2}\ ;\ 2\pi \ \Big]\\\\1-sin^2x=5+5sinx\ \ ,\ \ \ sin^2x+5sinx+4=0\ \ \Rightarrow \\\\(sinx)=-4\ ,\ (sinx)=-1\ \ \ (teorema Vieta)\\\\a)\ \ sinx=-4\ \ \to \ \ \ x\in \varnothing\ \ \ \ \ (\ |sinx|\leq 1\ )\\\\b)\ \ sinx=-1\ \ ,\ \ x=-\dfrac{\pi}{2}+2\pi n\ ,\ n\in Z\\\\c)\ \ x\in \Big[-\dfrac{\pi }{2}\ ;\ 2\pi \ \Big]\ :\ \ \ x_1=-\dfrac{\pi}{2}\ ,\ \ x_2=\dfrac{3\pi }{2}\\\\\\-\dfrac{\pi}{2}+\dfrac{3\pi}{2}=\dfrac{2\pi}{2}=\boxed{\pi }$

$Otvet:\ \ \pi \ .$

$24)\ \ \ y=arccos\dfrac{1}{2x-5}\\\\\\OOF:\ \ \ -1\leq \dfrac{1}{2x-5}\leq 1\ \ \ \to \ \ \ \ \left\{\begin{array}{l}\dfrac{1}{2x-5}\leq 1\\\dfrac{1}{2x-5}\geq -1\end{array}\right\ \ \left\{\begin{array}{l}\dfrac{1}{2x-5}-1\leq 0\\\dfrac{1}{2x-5}+1\geq 0\end{array}\right$

$\left\{\begin{array}{l}\dfrac{1-2x+5}{2x-5}\leq 0\\\\\dfrac{1+2x-5}{2x-5}\geq 0\end{array}\right\ \ \left\{\begin{array}{l}\dfrac{6-2x}{2x-5}\leq 0\\\ \dfrac{2x-4}{2x-5}\geq 0 \end{array}\right\ \ \left\{\begin{array}{l}\dfrac{-2(x-3)}{2x-5}\leq 0\\\dfrac{2(x-2)}{2x-5}\geq 0\end{array}\right$

$\left\{\begin{array}{l}\dfrac{x-3}{2x-5}\geq 0\\\ \dfrac{x-2}{2x-5}\leq 0\end{array}\right\ \ \left\{\begin{array}{l}x\in (-\infty ;\ \dfrac{5}{2}\ )\cup [\ 3\ ;+\infty \, )\\x\in (-\infty \, ;\ 2\ ]\cup (\ \dfrac{5}{2}\ ;+\infty \, )\end{array}\right\ \ \ \ \Rightarrow \\\\\\\boxed{x\in (-\infty \, ;\ 2\ ]\cup [\ 3\ ;+\infty \, )\ }$

Ответ:  Б .