\(\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2x^2}{x^2-4}\left(x\ne\pm2\right)\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{2x^2}{x^2-4}\)
\(\Rightarrow x^2+3x+2+x^2-3x+2=2x^2\)
\(\Leftrightarrow0x=-4\left(v\text{ô}nghi\text{ệm}\right)\)
Vậy phương trình vô nghiệm