Question

спочнорзррщозозозозозррщ​

Answer 1

Ответ:

$4)\ \ \Big(\dfrac{1}{2}\Big)^{x-2}>\dfrac{1}{8}\ \ \ \to \ \ \ \Big(\dfrac{1}{2}\Big)^{x-2}>\Big(\dfrac{1}{2}\Big)^3\\\\0<\dfrac{1}{2}<1\ \ \Rightarrow \ \ \ y=\Big(\dfrac{1}{2}\Big)^{x}\ -\ ybuvayuchaya\ \ \to \ \ x-2<3\ \ ,\ \ x<5\\\\x\in (-\infty ;\, 5\, )$

$5)\ \ f(x)\ vozrasnaet:\ \ x\in [-3\ ;\ 0\ ]\\\\\\6)\ \ 2cos^2\dfrac{a}{2}-cosa-1=2cos^2\dfrac{a}{2}-(1+cosa)=2cos^2\dfrac{a}{2}-2cos^2\dfrac{a}{2}=0\\\\\\\star \ \ cos^2x=\dfrac{1+cos2x}{2}\ \ \to \ \ 1+cos2x=2cos^2x\ \ \to \ \ 1+cosa=2cos^2\dfrac{a}{2}\ \ \star$

$7)\ \ log_2x=\dfrac{1}{2}\ \ ,\ \ \ \ ODZ:\ x>0\ ,\\\\x=\Big(\dfrac{1}{2}\Big)^2\ \ ,\ \ x=\dfrac{1}{4}\ -\ otvet$