ĐKXĐ:
\(x\ne\pm2\)
\(\frac{x+1}{x-2}-\frac{x-1}{x+2}=\frac{x\left(x+4\right)}{x^2-4}\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)-\left(x-1\right)\left(x-2\right)}{x^2-4}=\frac{x\left(x+4\right)}{x^2-4}\)
\(\Leftrightarrow x^2+3x+2-x^2+3x-2=x^2+4x\)
\(\Leftrightarrow x^2+4x-6x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(chọn\right)\\x=2\left(loại\right)\end{matrix}\right.\)
KL:...................
\(\frac{x+1}{x-2}-\frac{x-1}{x+2}=\frac{x\left(x+4\right)}{x^2-4}\)
ĐKXĐ: x ≠ 2; x ≠ -2
⇒\(\frac{\left(x+1\right)\left(x+2\right)-\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{x\left(x+4\right)}{\left(x-2\right)\left(x+2\right)}\)
⇔\(x^2+3x+2-x^2+3x-2=x^2+4x\)
⇔\(6x=x^2+4x\)
⇔\(-x^2-4x+6x=0\)
⇔\(-x^2+2x=0\)
⇔\(-x\left(x-2\right)\)\(=0\)
⇒ -x = 0
⇔x = 0(thỏa mãn)
Hoặc x - 2 = 0
⇔x = 2(không thỏa mãn)
Vậy nghiệm của PT là x = 0