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