Đáp án:
Đặt
$\begin{array}{l}
\dfrac{a}{b} = \dfrac{c}{d} = k\\
\Rightarrow \left\{ \begin{array}{l}
a = b.k\\
c = d.k
\end{array} \right.\\
+ )\dfrac{{5{a^2} + 7ab}}{{3{a^2} - 4{b^2}}} = \dfrac{{5{{\left( {b.k} \right)}^2} + 7.b.k.b}}{{3{{\left( {b.k} \right)}^2} - 4{b^2}}}\\
= \dfrac{{{b^2}\left( {5{k^2} + 7k} \right)}}{{{b^2}\left( {3{k^2} - 4} \right)}} = \dfrac{{5{k^2} + 7k}}{{3{k^2} - 4}}\\
+ )\dfrac{{5{c^2} + 7cd}}{{3{c^2} - 4{d^2}}} = \dfrac{{5{{\left( {d.k} \right)}^2} + 7.d.k.d}}{{3{{\left( {d.k} \right)}^2} - 4{d^2}}}\\
= \dfrac{{{d^2}\left( {5{k^2} + 7k} \right)}}{{{d^2}\left( {3{k^2} - 4} \right)}} = \dfrac{{5{k^2} + 7k}}{{3{k^2} - 4}}\\
\Rightarrow \dfrac{{5{a^2} + 7ab}}{{3{a^2} - 4{b^2}}} = \dfrac{{5{c^2} + 7cd}}{{3{c^2} - 4{d^2}}}
\end{array}$
