Question

Phép trừ

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

Answer 1

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

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

\(=\dfrac{2x^4-6x^3+6x^2-6-2x+6}{\left(x^2-1\right)\left(x-3\right)}\)

\(=\dfrac{2x^4-6x^3+6x^2-2x}{\left(x^2-1\right)\left(x-3\right)}\)