Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
*)\\
\dfrac{{{x^3} - 6{x^2} + 9x}}{{4{x^2} - 36}} = \dfrac{{x.\left( {{x^2} - 6x + 9} \right)}}{{4.\left( {{x^2} - 9} \right)}} = \dfrac{{x.\left( {{x^2} - 2.x.3 + {3^2}} \right)}}{{4.\left( {{x^2} - {3^2}} \right)}} = \dfrac{{x.{{\left( {x - 3} \right)}^2}}}{{4.\left( {x - 3} \right)\left( {x + 3} \right)}} = \dfrac{{x.\left( {x - 3} \right)}}{{4.\left( {x + 3} \right)}}\\
*)\\
\dfrac{{2{x^3} - 8x}}{{{x^2} - 3x + 2}} = \dfrac{{2x.\left( {{x^2} - 4} \right)}}{{\left( {{x^2} - x} \right) + \left( { - 2x + 2} \right)}} = \dfrac{{2x.\left( {{x^2} - {2^2}} \right)}}{{x.\left( {x - 1} \right) - 2.\left( {x - 1} \right)}} = \dfrac{{2x.\left( {x - 2} \right)\left( {x + 2} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \dfrac{{2x.\left( {x + 2} \right)}}{{x - 1}}\\
*)\\
\dfrac{{{a^2} + ab}}{{2{a^2} + ab - {b^2}}} = \dfrac{{a.\left( {a + b} \right)}}{{\left( {2{a^2} + 2ab} \right) + \left( { - ab - {b^2}} \right)}} = \dfrac{{a.\left( {a + b} \right)}}{{2a.\left( {a + b} \right) - b.\left( {a + b} \right)}} = \dfrac{{a\left( {a + b} \right)}}{{\left( {a + b} \right)\left( {2a - b} \right)}} = \dfrac{a}{{2a - b}}
\end{array}\)
Em xem lại đề câu 1 nhé!
