Question

Rút gọn phân thức \(\frac{x^3-7x-6}{x^2\left(x-3\right)^2+4x\left(x-3\right)^2+4\left(x-3\right)^2}\)

Answer 1

\(\frac{x^3-7x-6}{x^2\left(x-3\right)^2+4x\left(x-3\right)^2+4\left(x-3\right)^2}\)

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

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

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

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

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

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