Question

СРОЧНО , АЛГЕБРА 11 КЛАСС.

Answer 1

Ответ: [-1;2]

Объяснение: детали во вложении

Answer 2

Ответ:

$4\cdot 3^{2x+1}-35\cdot 6^{x}+18\cdot 4^{x}\leq 0\ \,\ \ \ \ ODZ:\ x\in R\ ,\\\\4\cdot 3\cdot (3^{x})^2-35\cdot 3^{x}\cdot 2^{x}+18\cdot (2^{x})^2\leq 0\ \Big|:(2^{x})^2>0\\\\\\12\cdot \Big(\Big(\dfrac{3}{2}\Big)^{x}\Big)^2-35\cdot \Big(\dfrac{3}{2}\Big)^{x}+18\leq 0\\\\\\t=\Big(\dfrac{3}{2}\Big)^{x}>0\ \ ,\ \ 12\, t^2-35\, t+18\leq 0\ \ ,\\\\12\, t^2-35\, t+18=0\ ,\ D=261=19^2\ ,\ x_1=\dfrac{16}{2\cdot 12}=\dfrac{2}{3}\ \ ,\ x_2=\dfrac{54}{2\cdot 12}=\dfrac{9}{4}$

$12\Big(t-\dfrac{2}{3}\Big)\Big(t-\dfrac{9}{4}\Big)\leq 0\\\\znaki:\ \ \ +++[\ \dfrac{2}{3}\ ]---[\ \dfrac{9}{4}\ ]+++\ \ \ \ \left\{\begin{array}{l}t\geq \dfrac{2}{3}\\\ t\leq \dfrac{9}{4}\end{array}\right\\\\1)\ \ \Big(\dfrac{3}{2}\Big)^{x}\geq \dfrac{2}{3}\ \ ,\ \ \ \Big(\dfrac{3}{2}\Big)^{x}\geq \Big(\dfrac{3}{2}\Big)^{-1}\\\\y=\Big(\dfrac{3}{2}\Big)^{x}\ -\ vozrastsyushaya\ \ \Rightarrow \ \ x\geq -1\ \ ,\ \ x\in [\, -1\ ;+\infty )$

$2)\ \ \Big(\dfrac{3}{2}\Big)^{x}\leq \dfrac{9}{4}\ \ ,\ \ \ \Big(\dfrac{3}{2}\Big)^{x}\leq \Big(\dfrac{3}{2}\Big)^{2}\\\\y=\Big(\dfrac{3}{2}\Big)^{x}\ -\ vozrastsyushaya\ \ \Rightarrow \ \ x\leq 2\ \ ,\ \ x\in (-\infty ;\ 2\ ]\\\\\\Otvet:\ \ x\in [\, -1\ ;\ 2\ ]\ .$