Answer
The simplified form of a complex rational expression $\frac{t+\frac{1}{t}}{t-\frac{2}{t}}$ is $\frac{{{t}^{2}}+1}{{{t}^{2}}-2}$.
Work Step by Step
$\frac{t+\frac{1}{t}}{t-\frac{2}{t}}$
The numerator and denominator of the complex rational expression are not single, so perform the operation and get a single expression.
The numerator LCD of $t+\frac{1}{t}$ is $t$, and the denominator LCD of $t-\frac{2}{t}$ is $t$.
Solve the above expression by making all fractions equivalent as their LCD:
$\begin{align}
& \frac{t+\frac{1}{t}}{t-\frac{2}{t}}=\frac{t\cdot \frac{t}{t}+\frac{1}{t}}{t\cdot \frac{t}{t}-\frac{2}{t}} \\
& =\frac{\frac{{{t}^{2}}}{t}+\frac{1}{t}}{\frac{{{t}^{2}}}{t}-\frac{2}{t}} \\
& =\frac{\frac{{{t}^{2}}+1}{t}}{\frac{{{t}^{2}}-2}{t}}
\end{align}$
Divide the numerator by the denominator,
$\frac{t+\frac{1}{t}}{t-\frac{2}{t}}=\frac{{{t}^{2}}+1}{t}\div \frac{{{t}^{2}}-2}{t}$
$\begin{align}
& \frac{A}{B}\div \frac{C}{D}=\frac{A}{B}\cdot \frac{D}{C} \\
& =\frac{AD}{BC}
\end{align}$
where $\frac{A}{B},\left( B\ne 0 \right)$, $\frac{C}{D},\left( D\ne 0 \right)$ are rational expressions with $\frac{C}{D}\ne 0$
The reciprocal of $\frac{{{t}^{2}}-2}{t}$ is $\frac{t}{{{t}^{2}}-2}$
So, multiply the reciprocal of the divisor,
$\frac{t+\frac{1}{t}}{t-\frac{2}{t}}=\frac{{{t}^{2}}+1}{t}\cdot \frac{t}{{{t}^{2}}-2}$
Simplify the terms,
$\begin{align}
& \frac{t+\frac{1}{t}}{t-\frac{2}{t}}=\frac{\left( {{t}^{2}}+1 \right)}{t}\cdot \frac{t}{{{t}^{2}}-2} \\
& =\frac{{{t}^{2}}+1}{{{t}^{2}}-2}\cdot \frac{t}{t} \\
& =\frac{{{t}^{2}}+1}{{{t}^{2}}-2}
\end{align}$
