Question

СРОЧНО ПОМОГИТЕ РЕШИТЬ (

Answer 1

Ответ:7. D; 8. C; 9. A

Объяснение:

$7. \log_{0,3}\left(12-\dfrac{x}{2}\right)>1\\12-\dfrac{x}{2} \in (0; 0,3)\\-\dfrac{x}{2} \in (-12; -11,7)\\x \in (23,4; 24)$

$8. \: 4^{x-2}<\dfrac{\lg\sqrt{10}}{2}\\4^{x-2}<4^{-1}\\x-2<-1\\x<1\\x \in (-\infty; 1)$

$9. \cot x \cdot\sin x<0\\\\\dfrac{\cos x}{\sin x}\cdot\sin x<0\\\\\cos x <0 \Leftrightarrow 2 \: u \: 3$

Answer 2

$7)log_{0,3}(12-\frac{x}{2})>1\\\\ODZ:\\12-\frac{x}{2}>0\\\\-\frac{x}{2} >-12\\\\x<24\\\\0<0,3<1 \ \Rightarrow 12-\frac{x}{2}<0,3 \\\\-\frac{x}{2}<-11,7\\\\x>23,4\\\\\boxed{x\in(23,4; \ 24)}$

$8)4^{x-2}<\frac{lg\sqrt{10}}{2}\\\\4^{x-2}<\frac{lg10^{\frac{1}{2}}}{2}\\\\4^{x-2}<\frac{\frac{1}{2}lg10}{2}\\\\4^{x-2}<\frac{\frac{1}{2}}{2} \\\\4^{x-2}<\frac{1}{4}\\\\4^{x-2} <4^{-1}\\\\4>1 \ \Rightarrow x-2<-1\\\\x<1\\\\\boxed{x\in(-\infty;1)}\\\\\\9)Ctgx*Sinx=\frac{Cosx}{Sinx}*Sinx=Cosx$

Cosx < 0 во 2 и 3 координатных четвертях.