The function is $\"h(x)=\\frac{1}{\\sqrt[4]{x^2-5x}}\"$ .

All possible values of $\"x\"$ is the domain of the function.

The denominator of the function should not be zero.

So fourth root of the function should be grater than zero.

$\"x^2-5x>$

$\"x(x-5)>$

If $\"x\"$ is negative then the function becomes positive.

So it occurs in cases $\"x(x-5)>$ and $\"x(x-5)<$ .

Solve for $\"x(x-5)>$ .

$\"x>$ and $\"x>$

Solve for $\"x(x-5)<$ .

$\"x<$ and $\"x<$

Then the domain of the function is $\"x<$ and $\"x>$ .

The domain of the function is $\"(-\\infty,$ .