Question

помогите решить уравнение 11 класс

Answer 1

$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]