Đáp án:
$\begin{array}{l}
\dfrac{a}{b} = \dfrac{c}{d} = k \Leftrightarrow \left\{ \begin{array}{l}
a = b.k\\
c = d.k
\end{array} \right.\\
a)\dfrac{{a + c}}{c} = \dfrac{{bk + dk}}{{dk}}\\
= \dfrac{{\left( {b + d} \right).k}}{{dk}} = \dfrac{{b + d}}{d}\\
b)\dfrac{{3a + 5b}}{{3a - 5b}} = \dfrac{{3.bk + 5b}}{{3bk - 5b}} = \dfrac{{b\left( {3k + 5} \right)}}{{b\left( {3k - 5} \right)}} = \dfrac{{3k + 5}}{{3k - 5}}\\
\dfrac{{3c + 5d}}{{3c - 5d}} = \dfrac{{3dk + 5d}}{{3dk - 5d}} = \dfrac{{d\left( {3k + 5} \right)}}{{d\left( {3k - 5} \right)}} = \dfrac{{3k + 5}}{{3k - 5}}\\
\Leftrightarrow \dfrac{{3a + 5b}}{{3a - 5b}} = \dfrac{{3c + 5d}}{{3c - 5d}}\\
c)\dfrac{{ab}}{{cd}} = \dfrac{{bk.b}}{{dk.d}} = \dfrac{{{b^2}}}{{{d^2}}}\\
\dfrac{{{a^2} - {b^2}}}{{{c^2} - {d^2}}} = \dfrac{{{b^2}{k^2} - {b^2}}}{{{d^2}{k^2} - {d^2}}} = \dfrac{{{b^2}\left( {{k^2} - 1} \right)}}{{{d^2}\left( {{k^2} - 1} \right)}} = \dfrac{{{b^2}}}{{{d^2}}}\\
\Leftrightarrow \dfrac{{ab}}{{cd}} = \dfrac{{{a^2} - {b^2}}}{{{c^2} - {d^2}}}\\
d){\left( {\dfrac{{a - b}}{{c - d}}} \right)^2} = {\left( {\dfrac{{bk - b}}{{dk - d}}} \right)^2} = \dfrac{{{b^2}}}{{{d^2}}}\\
\dfrac{{ab}}{{cd}} = \dfrac{{{b^2}}}{{{d^2}}}\\
\Leftrightarrow {\left( {\dfrac{{a - b}}{{c - d}}} \right)^2} = \dfrac{{ab}}{{cd}}
\end{array}$
