Answer
$x\lt0$ or $x\gt5$
Work Step by Step
$\because h(x)$ is defined for all real x where the denominator is not equal to zero $[\sqrt[4] (x^{2}-5x)\ne0, x^{2}-5x\ne0$],
and only the fourth root of non-negative numbers is real $[x^{2}-5x\geq0]$
$\therefore x^{2}-5x\gt0,$
$x(x-5)\gt0,$
$x\lt0$ or $x\gt5$
