Giải thích các bước giải:
Ta có:
\[\begin{array}{l}
\frac{{3{x^3} - 12{x^2} - 15x}}{{6{x^3} + 9{x^2} + 3x}} = \frac{{3x\left( {{x^2} - 4x - 5} \right)}}{{3x\left( {2{x^2} + 3x + 1} \right)}}\\
= \frac{{{x^2} - 4x - 5}}{{2{x^2} + 3x + 1}} = \frac{{\left( {{x^2} + x} \right) - \left( {5x + 5} \right)}}{{\left( {2{x^2} + 2x} \right) + \left( {x + 1} \right)}} = \frac{{x\left( {x + 1} \right) - 5\left( {x + 1} \right)}}{{2x\left( {x + 1} \right) + \left( {x + 1} \right)}}\\
= \frac{{\left( {x + 1} \right)\left( {x - 5} \right)}}{{\left( {x + 1} \right)\left( {2x + 1} \right)}} = \frac{{x - 5}}{{2x + 1}}\\
\frac{{4{x^2} - 12x - 40}}{{8{x^2} + 20x + 8}} = \frac{{4\left( {{x^2} - 3x - 10} \right)}}{{4\left( {2{x^2} + 5x + 2} \right)}} = \frac{{{x^2} - 3x - 10}}{{2{x^2} + 5x + 2}}\\
= \frac{{\left( {{x^2} - 5x} \right) + \left( {2x - 10} \right)}}{{\left( {2{x^2} + 4x} \right) + \left( {x + 2} \right)}} = \frac{{\left( {x - 5} \right)\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {2x + 1} \right)}} = \frac{{x - 5}}{{2x + 1}}\\
\frac{{2{x^2} - 7x + 6}}{{{x^2} - 4}} = \frac{{\left( {2{x^2} - 4x} \right) - \left( {3x - 6} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \frac{{2x\left( {x - 2} \right) - 3\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \frac{{\left( {x - 2} \right)\left( {2x - 3} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \frac{{2x - 3}}{{x + 2}}\\
\frac{{{x^3} - {x^2} - x + 1}}{{{x^4} - 2{x^2} + 1}} = \frac{{\left( {{x^3} - {x^2}} \right) - \left( {x - 1} \right)}}{{{{\left( {{x^2} - 1} \right)}^2}}} = \frac{{{x^2}\left( {x - 1} \right) - \left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^2}{{\left( {x + 1} \right)}^2}}} = \frac{{\left( {x - 1} \right)\left( {{x^2} - 1} \right)}}{{{{\left( {x - 1} \right)}^2}{{\left( {x + 1} \right)}^2}}} = \frac{1}{{x + 1}}
\end{array}\]
