Answer
The simplified form of a complex rational expression $\frac{x-\frac{3}{x-2}}{x-\frac{12}{x+1}}$ is $\frac{{{\left( x+1 \right)}^{2}}}{\left( x-2 \right)\left( x+4 \right)}$.
Work Step by Step
$\frac{x-\frac{3}{x-2}}{x-\frac{12}{x+1}}$
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 $x-\frac{3}{x-2}$ is $\left( x-2 \right)$, and the denominator LCD of $x-\frac{12}{x+1}$ is $\left( x+1 \right)$.
Solve the expression by making all fractions equivalent to their LCD:
$\begin{align}
& \frac{x-\frac{3}{x-2}}{x-\frac{12}{x+1}}=\frac{x\cdot \frac{\left( x-2 \right)}{\left( x-2 \right)}-\frac{3}{x-2}}{x\cdot \frac{\left( x+1 \right)}{\left( x+1 \right)}-\frac{12}{x+1}} \\
& =\frac{\frac{x\left( x-2 \right)}{x-2}-\frac{3}{\left( x-2 \right)}}{\frac{x\left( x+1 \right)}{\left( x+1 \right)}-\frac{12}{\left( x+1 \right)}} \\
& =\frac{\frac{x\left( x-2 \right)-3}{\left( x-2 \right)}}{\frac{x\left( x+1 \right)-12}{\left( x+1 \right)}}
\end{align}$
Divide the numerator by the denominator,
$\frac{x-\frac{3}{x-2}}{x-\frac{12}{x+1}}=\frac{x\left( x-2 \right)-3}{\left( x-2 \right)}\div \frac{x\left( x+1 \right)-12}{\left( x+1 \right)}$
$\begin{align}
& \frac{A}{B}\div \frac{C}{D}=\frac{A}{B}\cdot \frac{D}{C} \\
& =\frac{AD}{BC}
\end{align}$
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 $\frac{x\left( x+1 \right)-12}{\left( x+1 \right)}$ is $\frac{\left( x+1 \right)}{x\left( x+1 \right)-12}$
So, multiply the reciprocal of the divisor,
$\frac{x-\frac{3}{x-2}}{x-\frac{12}{x+1}}=\frac{x\left( x-2 \right)-3}{\left( x-2 \right)}\cdot \frac{\left( x+1 \right)}{x\left( x+1 \right)-12}$
Simplify the terms,
$\begin{align}
& \frac{x-\frac{3}{x-2}}{x-\frac{12}{x+1}}=\frac{x\left( x-2 \right)-3}{\left( x-2 \right)}\cdot \frac{\left( x+1 \right)}{x\left( x+1 \right)-12} \\
& =\frac{{{x}^{2}}-2x-3}{\left( x-2 \right)}\cdot \frac{\left( x+1 \right)}{{{x}^{2}}+x-12} \\
& =\frac{\left( {{x}^{2}}-2x-3 \right)\left( x+1 \right)}{\left( x-2 \right)\left( {{x}^{2}}+x-12 \right)}
\end{align}$
$\begin{align}
& \left( {{x}^{2}}-2x-3 \right)\left( x+1 \right)=\left( {{x}^{2}}+x-3x-3 \right)\left( x+1 \right) \\
& =\left( x\left( x+1 \right)-3\left( x+1 \right) \right)\left( x+1 \right) \\
& =\left( x+1 \right)\left( x-3 \right)\left( x+1 \right)
\end{align}$
$\begin{align}
& \left( x-2 \right)\left( {{x}^{2}}+x-12 \right)=\left( x-2 \right)\left( {{x}^{2}}+4x-3x-12 \right) \\
& =\left( x-2 \right)\left( x\left( x+4 \right)-3\left( x+4 \right) \right) \\
& =\left( x-2 \right)\left( x+4 \right)\left( x-3 \right)
\end{align}$
Thus, the expression becomes:
$\begin{align}
& \frac{x-\frac{3}{x-2}}{x-\frac{12}{x+1}}=\frac{\left( x+1 \right)\left( x-3 \right)\left( x+1 \right)}{\left( x-2 \right)\left( x+4 \right)\left( x-3 \right)} \\
& =\frac{\left( x+1 \right)\left( x+1 \right)\left( x-3 \right)}{\left( x-2 \right)\left( x+4 \right)\left( x-3 \right)} \\
& =\frac{{{\left( x+1 \right)}^{2}}}{\left( x-2 \right)\left( x+4 \right)}
\end{align}$
