Question

ПОМОГИТЕ ПОЖАЛУЙСТА!!!

Answer 1

Ответ:

$-3;-2;-\sqrt{2};~\sqrt{2};~2;~3$

Объяснение:

$\frac{6}{(x^2-3)^2}=\frac{7}{|x^2-3|}-1 \\ \\ \frac{6}{|x^2-3|^2}=\frac{7}{|x^2-3|}-1 \\ \\ |x^2-3|=t, ~~~t\neq 0 \\ \\ \frac{6}{t^2}=\frac{7}{t}-1 \\ \\ 1-\frac{7}{t}+\frac{6}{t^2}=0~~~~~~|\cdot t^2\neq 0 \\ \\ t^2-7t+6=0$

по теореме Виета  $t_1+t_2=7,~~t_1\cdot t_2=6$

$\left[\begin{array}{c}{t_1=1}&{t_2=6}\end{array}$

$\left[\begin{array}{c}{|x^2-3|=1}&{|x^2-3|=6}\end{array}$

$\left[\begin{array}{c}{x^2-3=-1}&{x^2-3=1}&{x^2-3=-6}\\{x^2-3=6}\end{array}$

$\left[\begin{array}{c}{x^2=2}&{x^2=4}&{x^2=-3}\\{x^2=9}\end{array}$

$\left[\begin{array}{c}{x=\pm \sqrt{2}}&{x=\pm \sqrt{4}}&{x\in \emptyset}\\{x= \pm \sqrt{9}}\end{array}$

$\left[\begin{array}{c}{x=\pm \sqrt{2}}&{x=\pm 2}&{x= \pm 3}\end{array}$