Answer
$\displaystyle \frac{(1-x)^{2}}{x^{3/2}}$
Work Step by Step
$x^{-\frac{1}{2}}=x^{-\frac{3}{2}+1}=x\cdot x^{-\frac{3}{2}}$,
$x^{\frac{1}{2}}=x^{-\frac{3}{2}+\frac{4}{2}}=x^{2}\cdot x^{-\frac{3}{2}}$
So, we factor out $x^{-3/2}$
$= x^{-3/2}(1-2x+x^{2})$
the parentheses hold a square of a difference, $ (a -b)^{2}=a^{2}-2ab+b^{2}$.
$=x^{-3/2}\cdot(1-x)^{2}\qquad $... apply $a^{-n}=\displaystyle \frac{1}{a^{n}}$
= $\displaystyle \frac{(1-x)^{2}}{x^{3/2}}$
