Question

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

Answer 1

\(\frac{x-1}{x}+\frac{1}{x+1}=\frac{2x-1}{2x^2+2}\)\(\left(x\ne0,x\ne-1\right)\)

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

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

\(\Leftrightarrow2x^4-x^2+3x-2=0\)

Pt nghiệm vô tỉ