Решите неравенство .........................

$log_{x^2} (x+2) \leq 1$
ОДЗ:
$\left \{ {{x+2\ \textgreater \ 0} \atop {x^2\ \textgreater \ 0}}\atop {x^2 \neq 1}} \right.$
$\left \{ {{x\ \textgreater \ -2} \atop {x \neq 0}}\atop {x \neqб 1}} \right.$

----------(-2)----------(-1)--------(0)---------(1)-------------
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$x$ ∈ $(-2;-1)$ ∪ $(-1;0)$ ∪ $(0;1)$ ∪ $(1;+$ ∞ $)$

$log_{x^2} (x+2) \leq log_{x^2} x^2$
$\left \{ {{0\ \textless \ x^2\ \textless \ 1} \atop {x+2 \geq x^2}} \right.$ или $\left \{ {{x^2 \geq 1} \atop {x+2 \leq x^2}} \right.$
$\left \{ {{-1\ \textless \ x\ \textless \ 1} \atop {x \neq 0}} \atop {x^2-x-2 \leq 0}} \right.$ или $\left \{ { \left \ [ {{x\ \textless \ -1} \atop {x\ \textgreater \ 1}} \right. \atop {x^2-x-2 \geq 0}} \right.$
$x^{2} -x-2=0$
$D=1+8=9$
$x_1= \frac{1+3}{2}=2$
$x_1= \frac{1-3}{2}=-1$

$\left \{ {{-1\ \textless \ x\ \textless \ 1} \atop {x \neq 0}} \atop {(x-2)(x+1) \leq 0}} \right.$ или $\left \{ { \left \ [ {{x\ \textless \ -1} \atop {x\ \textgreater \ 1}} \right. \atop {(x-2)(x+1) \geq 0}} \right.$

-------[-1]-----------------[2]--------- ------------[-1]--------[2]----------
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------(-1)---(0)---(1)---------------- -------------(-1)----(1)-------------
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Ответ: (-2;-1) ∪ (-1;0) ∪ (0;1) ∪ [2; +∞)