{x2 > 0 ⇒ x ≠ 0
{(x–1)2 > 0 ⇒ x ≠ 1
logx2 (x–1)2 ≤ 1
logx2 (x–1)2 ≤ logx2 (x2)
Применяем метод рационализации логарифмических неравенств
(x2–1)·((x–1)2–x2) ≤ 0
(x–1)(x+1)((x–1)–x)·((x–1)+x)≤ 0
(x–1)(x+1)(–1)·(2x–1) ≤ 0
(x–1)(x+1)(2x–1) ≥ 0
_–__ [–1] ________+______[1/2] __–__ [1] __+__
С учетом ОДЗ
О т в е т. [–1;0) U(0;(1/2)] U(1;+ ∞ )