\(\left(\left(x+1\right)\left(2-x\right)-\left(x^2-1\right)\right):\left(x+1\right)=2\)
\(\Leftrightarrow\left(\left(x+1\right)\left(2-x\right)-\left(x-1\right)\left(x+1\right)\right):\left(x+1\right)=2\)
\(\Leftrightarrow\left(2-x\right)-\left(x-1\right)=2\)
\(\Leftrightarrow2-x-x+1=2\)
\(\Leftrightarrow-2x+1=0\Leftrightarrow x=\dfrac{1}{2}\)
\(\left[\left(x+1\right)\left(2-x\right)-\left(x^2-1\right)\right]:\left(x+1\right)=2\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(2-x\right)-\left(x-1\right)\left(x+1\right)}{x+1}=2\)
\(\Leftrightarrow2-x-x+1=2\)
\(\Leftrightarrow3-2x=2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
