Cho biểu thức: \(Q = \left( {\dfrac{{{x^2} - 1}}{{x - 1}} + \dfrac{{{x^3} - 1}}{{1 - {x^2}}}} \right):\dfrac{{2{x^2} - 4x + 2}}{{{x^2} - 1}}\quad \left( {x \ne  \pm {\rm{ }}1} \right)\).

Rút gọn Q ta được:

\(Q = \left( {\dfrac{{{x^2} - 1}}{{x - 1}} + \dfrac{{{x^3} - 1}}{{1 - {x^2}}}} \right):\dfrac{{2{x^2} - 4x + 2}}{{{x^2} - 1}}\quad \left( {x \ne  \pm 1} \right)\)

\(Q = \left( {\dfrac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{x - 1}} - \dfrac{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}} \right):\dfrac{{2{{\left( {x - 1} \right)}^2}}}{{{x^2} - 1}}\)

\(Q = \left( {\dfrac{{x + 1}}{1} - \dfrac{{{x^2} + x + 1}}{{x + 1}}} \right):\dfrac{{2\left( {x - 1} \right)}}{{x + 1}}\)

\(Q = \dfrac{{{{\left( {x + 1} \right)}^2} - {x^2} - x - 1}}{{x + 1}}.\dfrac{{x + 1}}{{2\left( {x - 1} \right)}}\)

\(\begin{array}{l}Q = \dfrac{{{x^2} + 2x + 1 - {x^2} - x - 1}}{{x + 1}}.\dfrac{{x + 1}}{{2\left( {x - 1} \right)}}\\Q = \dfrac{x}{{2\left( {x - 1} \right)}}\end{array}\)

Vậy \(Q = \dfrac{x}{{2\left( {x - 1} \right)}}\) với mọi \(x \ne  \pm 1\).