Answer
$\dfrac{t^{4}}{rs^{2}}$
Work Step by Step
Using the laws of exponents, the given expression, $
\left( \dfrac{r^{-1}s^2t^{-3}}{r^{-2}s^0t^1}\right)^{-1}
,$ is equivalent to
\begin{align*}
&
\dfrac{r^{-1(-1)}s^{2(-1)}t^{-3(-1)}}{r^{-2(-1)}s^{0(-1)}t^{1(-1)}}
&\left(\text{use }(a^x)^y=a^{xy}\right)
\\\\&=
\dfrac{r^{1}s^{-2}t^{3}}{r^{2}s^{0}t^{-1}}
\\\\&=
r^{1-2}s^{-2-0}t^{3-(-1)}
&\left(\text{use }\dfrac{a^x}{a^y}=a^{x-y}\right)
\\\\&=
r^{-1}s^{-2}t^{4}
\\\\&=
\dfrac{t^{4}}{r^{1}s^{2}}
&\left(\text{use }a^{-x}=\dfrac{1}{a^{x}}\right)
\\\\&=
\dfrac{t^{4}}{rs^{2}}
.\end{align*}
Hence, the simplified form of the given expression is $
\dfrac{t^{4}}{rs^{2}}
$.
