Question

Câu: Thực hiện phép tính

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

Answer 1

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

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

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

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

= \(\frac{5x+5}{\left(x+1\right).\left(x^2-x+1\right)}\)

= \(\frac{5\left(x+1\right)}{\left(x+1\right).\left(x^2-x+1\right)}\)

= \(\frac{5}{x^2-x+1}\)