Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

The simplified form of a complex rational expression $\frac{t+\frac{1}{t}}{t-\frac{2}{t}}$ is $\frac{{{t}^{2}}+1}{{{t}^{2}}-2}$.
$\frac{t+\frac{1}{t}}{t-\frac{2}{t}}$ 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}