Question

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

Answer 1

đề bài là j thế

Answer 2

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

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

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