Question

giúp với (thugon)

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

Answer 1

\(ĐK:x\ne1\\ A=\dfrac{2x^2+1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}:\dfrac{x^2+x+1-x^2+2}{x^2+x+1}\\ A=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+x+1}{x+3}\\ A=\dfrac{x}{x+3}\)

Answer 2

\(A=\left(\dfrac{2x^2+1}{x^3-1}-\dfrac{1}{x-1}\right):\left(1-\dfrac{x^2-2}{x^2+x+1}\right)\left(đk:x\ne1\right)\)

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

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