Question

с объяснениями
4,2
20 баллов ​

Answer 1

Ответ:

$2)\ \ \Big(\dfrac{2}{25}\Big)^{2x-1}\geq \Big(\dfrac{4}{625}\Big)^{3-x}\ \ \ \to \ \ \ \Big(\dfrac{2}{25}\Big)^{2x-1}\geq \Big(\dfrac{2}{25}\Big)^{6-2x}\\\\\\0<\dfrac{2}{25}<1\ \ \ \to \ \ \ 2x-1\leq 6-2x\ \ ,\ \ 4x\leq 7\ \ ,\ \ x\leq 1,75\\\\\\x\in (-\infty \, ;\ 1,75\ ]$

$4)\ \ 0,5\, \underbrace{ctgx\cdot tgx}_{1}-sin4x=0\ \ ,\ \ \ sin4x=0,5\ \ ,\\\\\\x=(-1)^{n}\cdot \dfrac{\pi}{6}+\pi n\ ,\ n\in Z$