(2-log2 (x+1))/(log2 (x-2))=1Помогите пожалуйста

$\frac{(2-log_{2}(x+1))}{log_{2}(x-2)} =1$
разберемся с ОДЗ:
$\left \{ {{x+1\ \textgreater \ 0} \atop {x-2\ \textgreater \ 0}} \right.$
$\left \{ {{x\ \textgreater \ -1} \atop {x\ \textgreater \ 2}} \right.$
$log_{2}(x-2) \neq 0$
$log_{2}(x-2) \neq log_{2}1$
$x-2 \neq 1$
$x \neq 3$
итак,
x∈(2;3)∪(3;+∞)

$2 = log_{2}4$
$\frac{log_{2}4-log_{2}(x+1) }{log_{2}(x-2)} =1 |*(log_{2}(x-2))$
$log_{2}4-log_{2}(x+1) = log_{2}(x-2)$
$log_{2} \frac{4}{x+1} = log_{2} x-2$
$\frac{4}{x+1} = x-2$
$4 = (x+1)(x-2)$
$4 = x^2-x-2$
$x^2-x-6=0$
$D = 1+24 = 5^2$
$x_{1} = \frac{1+5}{2} =3$ - не уд ОДЗ
$x_{2} = \frac{1-5}{2} =-2$ - не уд. ОДЗ

ответ: x∈∅