Question

Tính

\(\dfrac{x^2+x+6}{x^3-27}-\dfrac{1-x}{3x-x^2}\)

Answer 1

\(=\dfrac{x^2+x+6}{\left(x-3\right)\left(x^2+3x+9\right)}-\dfrac{x-1}{x\left(x-3\right)}\)

\(=\dfrac{x^3+x^2+6x-\left(x-1\right)\left(x^2+3x+9\right)}{x\left(x-3\right)\left(x^2+3x+9\right)}\)

\(=\dfrac{x^3+x^2+6x-\left(x^3+3x^2+9x-x^2-3x-9\right)}{x\left(x-3\right)\left(x^2+3x+9\right)}\)

\(=\dfrac{x^3+x^2+6x-x^3-2x^2-6x+9}{x\left(x-3\right)\left(x^2+3x+9\right)}\)

\(=\dfrac{-x^2+9}{x\left(x-3\right)\left(x^2+3x+9\right)}\)

\(=\dfrac{-\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)\left(x^2+3x+9\right)}=\dfrac{-\left(x+3\right)}{x\left(x^2+3x+9\right)}\)