Answer
The solution set of the equation $\frac{1}{{{x}^{2}}-3x+2}=\frac{1}{x+2}+\frac{5}{{{x}^{2}}-4}$ is $\left\{ \frac{-1\pm \sqrt{21}}{2} \right\}$.
Work Step by Step
The provided equation is $\frac{1}{{{x}^{2}}-3x+2}=\frac{1}{x+2}+\frac{5}{{{x}^{2}}-4}$.
Now factorize the expression ${{x}^{2}}-3x+2$ as,
$\begin{align}
& {{x}^{2}}-3x+2={{x}^{2}}-2x-x+2 \\
& =x\left( x-2 \right)-1\left( x-2 \right) \\
& =\left( x-2 \right)\left( x-1 \right)
\end{align}$
Use the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to simplify the expression ${{x}^{2}}-4$.
$\begin{align}
& {{x}^{2}}-4={{x}^{2}}-{{2}^{2}} \\
& =\left( x-2 \right)\left( x+2 \right)
\end{align}$
The provided equation $\frac{1}{{{x}^{2}}-3x+2}=\frac{1}{x+2}+\frac{5}{{{x}^{2}}-4}$ can be written as,
$\frac{1}{\left( x-2 \right)\left( x-1 \right)}=\frac{1}{x+2}+\frac{5}{\left( x-2 \right)\left( x+2 \right)}$
The least common denominator of the above equation is $\left( x-1 \right)\left( x-2 \right)\left( x+2 \right)$. Multiply the least common denominator by both sides.
$\begin{align}
& \left( x-1 \right)\left( x-2 \right)\left( x+2 \right)\left[ \frac{1}{\left( x-2 \right)\left( x-1 \right)} \right]=\left[ \frac{1}{x+2}+\frac{5}{\left( x-2 \right)\left( x+2 \right)} \right]\left( x-1 \right)\left( x-2 \right)\left( x+2 \right) \\
& \left( x+2 \right)=\left( x-1 \right)\left( x-2 \right)+5\left( x-1 \right)
\end{align}$
Use distributive property to simplify the above equation.
$\begin{align}
& x+2={{x}^{2}}-2x-x+2+5x-5 \\
& x+2={{x}^{2}}+2x-3
\end{align}$
Subtract $2$ from both sides.
$\begin{align}
& x+2-2={{x}^{2}}+2x-3-2 \\
& x={{x}^{2}}+2x-5
\end{align}$
Subtract $x$ from both sides.
$\begin{align}
& x-x={{x}^{2}}+2x-5-x \\
& {{x}^{2}}+x-5=0
\end{align}$
Use the formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to solve the ${{x}^{2}}+x-5=0$ equation.
Here, $a=1$, $b=1$ and $c=-5$.
The solution of the equation ${{x}^{2}}+x-5=0$ is,
$\begin{align}
& x=\frac{-1\pm \sqrt{{{1}^{2}}-4\left( 1 \right)\left( -5 \right)}}{2\left( 1 \right)} \\
& =\frac{-1\pm \sqrt{1+20}}{2} \\
& =\frac{-1\pm \sqrt{21}}{2}
\end{align}$
The solution set of the provided equation $\frac{1}{{{x}^{2}}-3x+2}=\frac{1}{x+2}+\frac{5}{{{x}^{2}}-4}$is $\left\{ \frac{-1\pm \sqrt{21}}{2} \right\}$.
