5^2/х меньше =0.2^х-3 решите

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