Question

1: rút gọn \(A=\dfrac{2}{x^2-1}-\dfrac{1}{x^2+x}+\dfrac{x^2-3}{x^3-x}\); \(B=\dfrac{2}{x-1}+\dfrac{2x-1}{x^2+x+1}+\dfrac{x^2+6x+2}{x^3-1}\)

2: tìm x: \(\dfrac{4}{3}\left(x-2\right)+\dfrac{\left(x-1\right)\left(x+2\right)}{2}=3-\dfrac{5x\left(1-2x\right)}{4}\)

Answer 1

1:

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

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

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

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

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