Question

Решить неравенство:

$log(_{x-1})^{2} x^{2} leq 1$

 

Answer 1

$log_{(x-1)^2}x^2 leq 1\ log_{(x-1)^2}frac{x^2}{(x-1)^2}leq0\ begin{cases} ((x-1)^2-1)left(frac{x^2}{(x-1)^2}-1right)le0\ xnotinlbrace0,1,2rbrace end{cases}\ begin{cases} x(x-2)(2x-1)le0\ xnotinlbrace0,1,2rbrace end{cases}\ xin(-infty,0)cup[0.5,1)cup(1,2)$