\(=\dfrac{x+1}{x\left(x-1\right)}-\dfrac{x+2}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+2x+1-x^2-2x-1}{x\left(x-1\right)\left(x+1\right)}=0\)
\(\dfrac{x+1}{x^2-x}+\dfrac{x+2}{1-x^2}-\dfrac{1}{x^3-x}\\ =\dfrac{x+1}{x\left(x-1\right)}-\dfrac{x+2}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x\left(x^2-1\right)}\\ =\dfrac{x+1}{x\left(x-1\right)}-\dfrac{x+2}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x+1\right)^2}{x\left(x-1\right)\left(x+1\right)}-\dfrac{x^2+2x}{x\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^2+2x+1-x^2-2x-1}{x\left(x-1\right)\left(x+1\right)}\\ =\dfrac{0}{x\left(x-1\right)\left(x+1\right)}\\ =0\)
giúp mình bài này với ạ,từng bước lun nhé,mình cảm ơn ạ
