Đáp án:
$\begin{array}{l}
a)\dfrac{{x - 2y}}{{{x^3} + {y^3}}}.A\left( x \right) = \dfrac{{{x^2} - 2xy}}{{{x^2} - xy + {y^2}}}\\
Do:\dfrac{{{x^2} - 2xy}}{{{x^2} - xy + {y^2}}}\\
= \dfrac{{x.\left( {x - 2y} \right)}}{{\left( {x + y} \right).\left( {{x^2} - xy + {y^2}} \right)}}.\left( {x + y} \right)\\
= \dfrac{{x.\left( {x + y} \right)\left( {x - 2y} \right)}}{{{x^3} + {y^3}}}\\
= x.\left( {x + y} \right).\dfrac{{x - 2y}}{{{x^3} + {y^3}}}\\
\Rightarrow A\left( x \right) = x.\left( {x + y} \right) = {x^2} + xy\\
b)\dfrac{{{x^2} + 3x}}{{x - 4}}:B\left( x \right) = \dfrac{{{x^2} - 9}}{{{x^2} - 4x}}\\
\Rightarrow \dfrac{{x\left( {x + 3} \right)}}{{\left( {x - 4} \right).B\left( x \right)}} = \dfrac{{\left( {x - 3} \right)\left( {x + 3} \right)}}{{x\left( {x - 4} \right)}}\\
\Rightarrow \dfrac{x}{{B\left( x \right)}}.\dfrac{{x + 3}}{{x - 4}} = \dfrac{{x - 3}}{x}.\dfrac{{x + 3}}{{x - 4}}\\
\Rightarrow \dfrac{x}{{B\left( x \right)}} = \dfrac{{x - 3}}{x}\\
\Rightarrow B\left( x \right) = \dfrac{{x - 3}}{{x.x}} = \dfrac{{x - 3}}{{{x^2}}}
\end{array}$
