Ta có: \(\dfrac{x+1}{2x-2}+\dfrac{2x}{1-x^2}\Rightarrow MTC=-2\left(x-1\right)\left(x+1\right)\)
Khi đó ta có:
\(\dfrac{x+1}{2x-2}+\dfrac{2x}{1-x^2}=\dfrac{-\left(x+1\right)^2}{-2\left(x-1\right)\left(x+1\right)}+\dfrac{2.2x}{-2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2-2x-1+4x}{2\left(1-x^2\right)}=\dfrac{-\left(x-1\right)^2}{-2\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{2\left(x+1\right)}\)
đk x khác -1 ; 1 ; - 2
\(\dfrac{\left(x+1\right)^2-4x}{2\left(x-1\right)\left(x+2\right)}=\dfrac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+2\right)}=\dfrac{x-1}{2\left(x+2\right)}\)
