5^2/х<=0.2^х-3 решите

$5^{ \frac{2}{x} } \leq 0,2^{x-3} \\ 5^{ \frac{2}{x} } \leq ( \frac{1}{5} )^{x-3} \\ 5^{ \frac{2}{x} } \leq 5^{-(x-3)} \\ \frac{2}{x} \leq 3-x \\ \frac{2}{x} - 3+x \leq 0 \\ \frac{ x^{2} -3x+2}{x} \leq 0$
$\left \{ {{x\ \textgreater \ 0} \atop {x^{2} -3x+2 \leq 0}} \right.$ или $\left \{ {{x\ \textless \ 0} \atop {x^{2} -3x+2 \geq 0}} \right.$
$D=9-8=1 \\ x_{1} = \frac{3-1}{2} =1, x_{2} =\frac{3+1}{2} =2$
$1) \left \{ {{x\ \textgreater \ 0} \atop {x^{2} -3x+2 \leq 0}} \right. \left \{ {{x\ \textgreater \ 0} \atop {1 \leq x \leq 2}} \right.$ ⇒ x∈[1,2]

$2) \left \{ {{x\ \textless \ 0} \atop {x^{2} -3x+2 \geq 0}} \right. \left \{ {{x\ \textless \ 0} \atop {x\ \textless \ 1, x\ \textgreater \ 2}} \right.$
⇒ x∈(-∞,0)
ответ: x∈(-∞,0)∨[1,2]