Question

5^2/х<=0.2^х-3 решите

Answer 1

$5^{ frac{2}{x} } leq 0,2^{x-3} \ 5^{ frac{2}{x} } leq ( frac{1}{5} )^{x-3} \ 5^{ frac{2}{x} } leq 5^{-(x-3)} \ frac{2}{x} leq 3-x \ frac{2}{x} - 3+x leq 0 \ frac{ x^{2} -3x+2}{x} leq 0$
$left { {{x textgreater 0} atop {x^{2} -3x+2 leq 0}} right.$ 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]

Answer 2

спасибо