#### Answer

$r-2+\dfrac{1}{r}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To simplify the given expression, $
(r^{1/2}-r^{-1/2})^2
,$ use the special product on squaring binomials. Then, use the laws of exponents to simplify the resulting expression.
$\bf{\text{Solution Details:}}$
Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(r^{1/2})^2-2(r^{1/2})(r^{-1/2})+(r^{-1/2})^2
.\end{array}
Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
r^{\frac{1}{2}\cdot2}-2(r^{1/2})(r^{-1/2})+r^{-\frac{1}{2}\cdot2}
\\\\=
r^{1}-2(r^{1/2})(r^{-1/2})+r^{-1}
\\\\=
r-2(r^{1/2})(r^{-1/2})+r^{-1}
.\end{array}
Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
r-2r^{\frac{1}{2}+\left(-\frac{1}{2}\right)}+r^{-1}
\\\\=
r-2r^{\frac{1}{2}-\frac{1}{2}}+r^{-1}
\\\\=
r-2r^{0}+r^{-1}
.\end{array}
Since any expression raised to the power of zero is $1$, the expression above is equivalent to
\begin{array}{l}\require{cancel}
r-2(1)+r^{-1}
\\\\=
r-2+r^{-1}
.\end{array}
Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
r-2+\dfrac{1}{r^1}
\\\\=
r-2+\dfrac{1}{r}
.\end{array}