Cho biểu thức

\(P = \frac{{2x - 6}}{{{x^3} - 3{x^2} - x + 3}} + \frac{{2{x^2}}}{{1 - {x^2}}} - \frac{6}{{x - 3}}\) (x ≠ 3, x ≠ 1, x ≠ –1).

Rút gọn phân thức \(\frac{{2x - 6}}{{{x^3} - 3{x^2} - x + 3}}\).

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

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