Answer
$\frac{-4t^2+5t+5}{t^2(t+1)}$
Work Step by Step
The $LCD$ of $t^2$ and $(t+1)$ is $t^2(t+1)$. Changing the given rational expressions to equivalent rational expressions that use the $LCD$, then
$$\begin{aligned}
&
\frac{5}{t^2}\cdot\frac{t+1}{t+1}-\frac{4}{t+1}\cdot\frac{t^2}{t^2}
\\&=
\frac{5t+5}{t^2(t+1)}-\frac{4t^2}{t^2(t+1)}
.\end{aligned}
$$
Adding/Subtracting similar fractions involves adding/subtracting the numerators and copying the common denominator. Therefore,
$$\begin{aligned}
\frac{5t+5}{t^2(t+1)}-\frac{4t^2}{t^2(t+1)}&=
\frac{5t+5-4t^2}{t^2(t+1)}
\\&=
\frac{-4t^2+5t+5}{t^2(t+1)}.
\end{aligned}
.$$Hence, the given expression simplifies to $\frac{-4t^2+5t+5}{t^2(t+1)}$.
