Question

Solve the following inequality:
${\left(\frac{1}{3}\right)}^{x+\frac{1}{2}-\frac{2}{x}}>\frac{1}{\sqrt{27}}$

Answer 1

${\left(\frac{1}{3}\right)}^{x+\frac{1}{2}-\frac{2}{x}}>\frac{1}{\sqrt{27}}$

$3^{\frac{2}{x}-x-\frac{1}{2}}>3^{\frac{-3}{2}}$

$\frac{2}{x}-x-\frac{1}{2}>\frac{-3}{2}$

$\frac{2}{x}-x+1>0$

$2-x^2+x<0$

$x^2-x-2<0$

$x^2-2x+x-2<0$

$(x-2)(x+1)<0$

$x\epsilon (-1,2)$