Đáp án:
$\begin{array}{l}
a)\dfrac{{{x^4} - x{y^3}}}{{2xy + {y^2}}}:\dfrac{{{x^3} + {x^2}y + x{y^2}}}{{2x + y}}\\
= \dfrac{{x\left( {{x^3} - {y^3}} \right)}}{{y\left( {2x + y} \right)}}.\dfrac{{2x + y}}{{x\left( {{x^2} + xy + {y^2}} \right)}}\\
= \dfrac{{\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)}}{{y\left( {{x^2} + xy + {y^2}} \right)}}\\
= \dfrac{{x - y}}{y}\\
b)\dfrac{{5{x^2} - 10xy + 5{y^2}}}{{2{x^2} - 2xy + 2{y^2}}}:\dfrac{{8x - 8y}}{{10{x^3} + 10{y^3}}}\\
= \dfrac{{5\left( {{x^2} - 2xy + {y^2}} \right)}}{{2\left( {{x^2} - xy + {y^2}} \right)}}.\dfrac{{10\left( {{x^3} + {y^3}} \right)}}{{8\left( {x - y} \right)}}\\
= \dfrac{{25.{{\left( {x - y} \right)}^2}.\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)}}{{8\left( {x - y} \right)\left( {{x^2} - xy + {y^2}} \right)}}\\
= \dfrac{{25\left( {x - y} \right)\left( {x + y} \right)}}{8}\\
= \dfrac{{25\left( {{x^2} - {y^2}} \right)}}{8}\\
39)\\
a)\dfrac{{{x^2} - 5x + 6}}{{{x^2} + 7x + 12}}:\dfrac{{{x^2} - 4x + 4}}{{{x^2} + 3x}}\\
= \dfrac{{\left( {x - 2} \right)\left( {x - 3} \right)}}{{\left( {x + 3} \right)\left( {x + 4} \right)}}.\dfrac{{x\left( {x + 3} \right)}}{{{{\left( {x - 2} \right)}^2}}}\\
= \dfrac{{x\left( {x - 3} \right)}}{{\left( {x + 4} \right)\left( {x - 2} \right)}}\\
= \dfrac{{{x^2} - 3x}}{{{x^2} + 2x - 8}}\\
b)\dfrac{{{x^2} + 2x - 3}}{{{x^2} + 3x - 10}}:\dfrac{{{x^2} + 7x + 12}}{{{x^2} - 9x + 14}}\\
= \dfrac{{\left( {x - 1} \right)\left( {x + 3} \right)}}{{\left( {x - 2} \right)\left( {x + 5} \right)}}.\dfrac{{\left( {x - 2} \right)\left( {x - 7} \right)}}{{\left( {x + 3} \right)\left( {x + 4} \right)}}\\
= \dfrac{{\left( {x - 1} \right)\left( {x - 7} \right)}}{{\left( {x + 5} \right)\left( {x + 4} \right)}}\\
= \dfrac{{{x^2} - 8x + 7}}{{{x^2} + 9x + 20}}
\end{array}$
