Question
Solve the following equation by factorization $\frac{x+1}{x-1}+\frac{x-2}{x+2}=3$
✓
Answer
$
\begin{aligned}
& \frac{x+1}{x-1}+\frac{x-2}{x+2}=3 \\
& \Rightarrow \frac{(x+1)(x+2)+(x-2)(x-1)}{(x-1)(x+2)}=3 \\
& \Rightarrow \frac{x^2+2 x+x+2+x^2-x-2 x+2}{x^2+2 x-x-2} \\
& \Rightarrow \frac{x^2+3 x+2+x^2-3 x+2}{x^2+x-2}=\frac{3}{1} \\
& \Rightarrow 2 x^2+4=3 x^2+3 x-6 \\
& \Rightarrow 2 x^2+4-3 x^2-3 x+6=0 \\
& \Rightarrow-x^2-3 x+10=0 \\
& \Rightarrow x^2+3 x-10=0 \\
& \Rightarrow x^2+5 x-2 x-10=0 \\
& \Rightarrow x(x+5)-2(x+5)=0 \\
& \Rightarrow(x+5)(x-2)=0
\end{aligned}
$
Either $x+5=0$,
then $x=-5$
or
$
x-2=0 \text {, }
$
then $x =2$
Hence $x=-5,2$.
\begin{aligned}
& \frac{x+1}{x-1}+\frac{x-2}{x+2}=3 \\
& \Rightarrow \frac{(x+1)(x+2)+(x-2)(x-1)}{(x-1)(x+2)}=3 \\
& \Rightarrow \frac{x^2+2 x+x+2+x^2-x-2 x+2}{x^2+2 x-x-2} \\
& \Rightarrow \frac{x^2+3 x+2+x^2-3 x+2}{x^2+x-2}=\frac{3}{1} \\
& \Rightarrow 2 x^2+4=3 x^2+3 x-6 \\
& \Rightarrow 2 x^2+4-3 x^2-3 x+6=0 \\
& \Rightarrow-x^2-3 x+10=0 \\
& \Rightarrow x^2+3 x-10=0 \\
& \Rightarrow x^2+5 x-2 x-10=0 \\
& \Rightarrow x(x+5)-2(x+5)=0 \\
& \Rightarrow(x+5)(x-2)=0
\end{aligned}
$
Either $x+5=0$,
then $x=-5$
or
$
x-2=0 \text {, }
$
then $x =2$
Hence $x=-5,2$.
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