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