Question

giải phương trình sau

\(\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)

Answer 1

ĐKXĐ: x\(\ne\pm2\)

\(\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{2\left(x^2+2\right)}{x^2-4}\)

\(\Rightarrow\left(x+1\right)\left(x+2\right)+\left(x-1\right)\left(x-2\right)=2\left(x^2+2\right)\)

\(\Leftrightarrow x^2+3x+2+x^2-3x+2=2\left(x^2+2\right)\)

\(\Leftrightarrow0x=0\)

Vậy: phương trình có vô số nghiệm (Với ĐKXĐ là x\(\ne\pm2\))