Question

Giải phương trình: \(\dfrac{1}{x-1}\)+\(\dfrac{1}{x-2}\)=\(\dfrac{1}{x+2}\)+\(\dfrac{1}{x+1}\)

Answer 1

ĐK : \(x\ne\pm1;x\ne\pm2\)

\(...\Leftrightarrow\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}\right)+\left(\dfrac{1}{x-2}-\dfrac{1}{x+2}\right)=0\)

\(\Leftrightarrow\dfrac{x+1-x+1}{x^2-1}+\dfrac{x+2-x+2}{x^2-4}=0\)

\(\Leftrightarrow\dfrac{2}{x^2-1}+\dfrac{4}{x^2-4}=0\)

\(\Leftrightarrow\dfrac{2x^2-8+4x^2-4}{\left(x^2-1\right)\left(x^2-4\right)}=0\)

\(\Leftrightarrow6x^2-12=0\)

\(\Rightarrow x^2=2\)

\(\Rightarrow x=\pm\sqrt{2}\) ( thỏa mãn)