Помогите решить уравнение 11 класс

Question 1

Question 2

$\frac{x}{x^2} = \frac{1}{x} \\ \\ 9 ^{\frac{1}{x} } -12*3 ^{\frac{1}{x} }+27 \leq 0 \\ \\( 3 ^{\frac{1}{x} })^2 -12*3 ^{\frac{1}{x} }+27 \leq 0$

Пусть
$3 ^{\frac{1}{x} }=t, \ t\ \textgreater \ 0$
тогда:

$t^2-12t+27 \leq 0 \\ \\ (t-3)(t-9) \leq 0 \\ \\ +++[3]---[9]+++\ \textgreater \ t \\ \\ t \in [3;9] \ \ \ \textless \ =\ \textgreater \ \left \{ {{t \geq 3} \atop {t \leq 9}} \right.$

обратная замена:
$\left \{ {{3^{ \frac{1}{x} }\geq 3^1} } \atop {{{3^{ \frac{1}{x} } \leq 3^2}} \right. \ \ \ \textless \ =\ \textgreater \ \ \ \left \{ {{ \frac{1}{x} \geq 1} \atop {\frac{1}{x} \leq 2}} \right. \ \ \ \textless \ =\ \textgreater \ \ \ \left \{ {{ \frac{1}{x} - 1 \geq 0} \atop {\frac{1}{x} - 2 \leq 0}} \right. \ \ \ \textless \ =\ \textgreater \ \ \left \{ {{ \frac{1-x}{x} \geq 0 } \atop {\frac{1-2x}{x} \leq 0}} \right. \\ \\ 1)\ \frac{1-x}{x} \geq 0 \\ \\ ---(0)+++[1]---\ \textgreater \ x \ \\ \\ 2) \ \frac{1-2x}{x} \leq 0 \\ \\ ---(0)+++[0.5]---\ \textgreater \ x \\ \\ x \in (-\infty; 0) \ U \ [0.5; +\infty)$

$\left \{ {{ \frac{1-x}{x} \geq 0 } \atop {\frac{1-2x}{x} \leq 0}} \right. \ \ \ \textless \ =\ \textgreater \ \ \ \left \{ {{x \in (0;\ 1]} \atop {x \in (-\infty; 0)\ U\ [0.5;+\infty)}} \right. \ \ \ \textless \ =\ \textgreater \ \ \ x \in [0.5;1]$

Ответ: х∈ [0.5; 1]