Question

[Trash Như]

Thực hiện phép tính:

\(\dfrac{3}{x}-\dfrac{x}{x-1}-\dfrac{x^2}{x+1}-\dfrac{x^2-3}{x^3-x}\)

Answer 1

[Nguyễn Hoàng Minh]

Nhận thấy \(x^3-x=x\left(x^2-1\right)=x\left(x-1\right)\left(x+1\right)\)

\(\dfrac{3}{x}-\dfrac{x}{x-1}-\dfrac{x^2}{x+1}-\dfrac{x^2-3}{x^3-x}\\ =\dfrac{3x^2-3-x^3-x^2-x^4+x^3-x^2+3}{x\left(x-1\right)\left(x+1\right)}\\ =\dfrac{-x^4+x^2}{x\left(x-1\right)\left(x+1\right)}=\dfrac{-x^2\left(x-1\right)\left(x+1\right)}{x\left(x-1\right)\left(x+1\right)}=-x\)