Log2 x+log2 (x-1)=<1

$log_{2}x+ log_{2}(x-1) \leq 1$

ОДЗ:
$\left \{ {{x\ \textgreater \ 0} \atop {x-1\ \textgreater \ 0}} \right.$

$\left \{ {{x\ \textgreater \ 0} \atop {x\ \textgreater \ 1}} \right.$

$x$ ∈ $(1;+$ ∞ $)$

$log_{2}x+ log_{2}(x-1) \leq log_{2}2$

$log_{2}(x^2-x)\leq log_{2}2$

$x^2-x\leq 2$

$x^2-x-2\leq 0$

$D=(-1)^2-4*1*(-2)=1+8=9$

$x_1= \frac{1+3}{2}=2$

$x_2= \frac{1-3}{2}=-1$

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$x$ ∈ $[-1;2]$

с учётом ОДЗ получаем

Ответ: $(1;2]$