\(=\dfrac{x^2+x+2x-2-x^2-1}{\left(x-1\right)\left(x+1\right)}=\dfrac{3x-3}{\left(x-1\right)\left(x+1\right)}=\dfrac{3}{x+1}\)
\(\dfrac{x}{x-1}+\dfrac{2}{x+1}-\dfrac{x^2+1}{x^2-1}\) đề ntn phải ko ạ?
\(=\dfrac{x}{x-1}+\dfrac{2}{x+1}-\dfrac{x^2+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+x}{\left(x-1\right)\left(x+1\right)}+\dfrac{2x-2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x^2+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+x+2x-2-x^2-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2-x^2+x+2x-2-1}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{3x-3}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{3}{x+1}\)
