Question

Решите два неравенства.

Answer 1

Ответ:

Объяснение:

$\displaystyle\bf\\a)\ \bigg(\frac{1}{3} \bigg)^{5x^2+8x-4}\leq 1;\ \bigg(\frac{1}{3} \bigg)^{5x^2+8x-4}\leq \bigg(\frac{1}{3} \bigg)^0\\\\\frac{1}{3} <1;\ \ \ 5x^2+8x-4\geq 0\\\\D=64-4\cdot5\cdot(-4)=144=12^2\\\\x_1=\frac{-8-12}{10}=-2 ;\ \ x_2=\frac{-8+12}{10}=0,4\\\\5(x+2)(x-0,4)\geq 0\\\\+++[-2]---[0,4]+++>x\\\\Otvet:\ x\in(-\infty;-2]\cup[0,4;+\infty)\\\\\\b)\ \bigg(\frac{4}{3} \bigg)^{2x-1}\geq \frac{3}{4} ;\ \bigg(\frac{4}{3} \bigg)^{2x-1}\geq\bigg(\frac{4}{3} \bigg)^{-1}$

$\displaystyle\bf\\\frac{4}{3} >1;\ \ 2x-1\geq -1;\ \ 2x\geq 0;\ \ x\geq 0\\\\Otvet:\ x\in[0;+\infty)$