Question

[Thanh Tú Võ]

giúp mình với

\(\frac{1}{x-1}-\frac{x^3-x}{x^2+1}\left(\frac{1}{x^2-2x+1}+\frac{1}{1-x^2}\right)\)

Answer 1

[Nguyễn Lê Phước Thịnh]

\(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\left(\dfrac{1}{\left(x-1\right)^2}-\dfrac{1}{\left(x-1\right)\left(x+1\right)}\right)\)

\(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\dfrac{x+1-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\)

\(=\dfrac{1}{x-1}-\dfrac{x\cdot2}{\left(x-1\right)\left(x^2+1\right)}\)

\(=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)