Решите уравнение методом замены переменной: (х²-2х)²-7(х²-2х)-8=0

$\displaystyle (x^2-2x)^2-7(x^2-2x)-8=0\\\\t=x^2-2x\\\\t^2-7t-8=0\\\\D=7^2+4\cdot8=49+32=81=9^2\\\\t_1=\frac{7+9}{2}=\frac{16}2=8\\\\t_2=\frac{7-9}2=-\frac{2}2=-1$

Обратная подстановка:

$x^2-2x=t, \quad t=8\\\\x^2-2x=8\\\\x^2-2x-8=0\\\\x^2-2x+1-9=0\\\\(x-1)^2-9=0\\\\(x-1+3)(x-1-3)=0\\\\(x+2)(x-4)=0\\\\x+2=0\quad\quad x-4=0\\\\\boxed{x=-2}\quad \quad \boxed{x=4}\\\\\\x^2-2x=t, \quad t=-1\\\\x^2-2x=-1\\\\x^2-2x+1=0\\\\(x-1)^2=0\\\\x-1=0\\\\\boxed{x=1}$