Đáp án:

$\begin{array}{l}
a)\dfrac{a}{b} = \dfrac{c}{d} = k \Leftrightarrow \left\{ \begin{array}{l}
a = b.k\\
c = d.k
\end{array} \right.\\
\dfrac{{2a + 3b}}{{2a - 3b}} = \dfrac{{2.b.k + 3.b}}{{2.b.k - 3b}} = \dfrac{{2k + 3}}{{2k - 3}}\\
\dfrac{{2c + 3d}}{{2c - 3d}} = \dfrac{{2.d.k + 3.d}}{{2.d.k - 3.d}} = \dfrac{{2k + 3}}{{2k - 3}}\\
 \Leftrightarrow \dfrac{{2a + 3b}}{{2a - 3b}} = \dfrac{{2c + 3d}}{{2c - 3d}}\\
2)\dfrac{{ab}}{{cd}} = \dfrac{{b.b.k}}{{d.d.k}} = \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}}}{{{d^2}}}\\
 \Leftrightarrow \dfrac{{ab}}{{cd}} = \dfrac{{{a^2} - {b^2}}}{{{c^2} - {d^2}}}\\
3)\dfrac{{7a - 4b}}{{3a + 5b}} = \dfrac{{7.b.k - 4b}}{{3.b.k + 5b}} = \dfrac{{b.\left( {7k - 4} \right)}}{{b\left( {3k + 5} \right)}} = \dfrac{{7k - 4}}{{3k + 5}}\\
\dfrac{{7c - 4d}}{{3c + 5d}} = \dfrac{{7.d.k - 4d}}{{3.d.k + 5d}} = \dfrac{{7k - 4}}{{3k + 5}}\\
 \Leftrightarrow \dfrac{{7a - 4b}}{{3a + 5b}} = \dfrac{{7c - 4d}}{{3c + 5d}}\\
4)\dfrac{{ac}}{{bd}} = \dfrac{{b.k.d.k}}{{b.d}} = {k^2}\\
\dfrac{{{a^2} + {c^2}}}{{{b^2} + {d^2}}} = \dfrac{{{b^2}{k^2} + {d^2}{k^2}}}{{{b^2} + {d^2}}} = {k^2}\\
 \Leftrightarrow \dfrac{{ac}}{{bd}} = \dfrac{{{a^2} + {c^2}}}{{{b^2} + {d^2}}}\\
5)\dfrac{{{a^3} + {b^3}}}{{{c^3} + {d^3}}} = \dfrac{{{b^3}{k^3} + {b^3}}}{{{d^3}{k^3} + {d^3}}} = \dfrac{{{b^3}}}{{{d^3}}}\\
\dfrac{{{{\left( {a + b} \right)}^3}}}{{{{\left( {c + d} \right)}^3}}} = \dfrac{{{b^3}}}{{{d^3}}}\\
 \Leftrightarrow \dfrac{{{a^3} + {b^3}}}{{{c^3} + {d^3}}} = \dfrac{{{{\left( {a + b} \right)}^3}}}{{{{\left( {c + d} \right)}^3}}}\\
b)\dfrac{{2a + 13b}}{{3a - 7b}} = \dfrac{{2c + 13d}}{{3c - 7d}}\\
G/s:\dfrac{a}{b} = \dfrac{c}{d} = k \Leftrightarrow \left\{ \begin{array}{l}
a = b.k\\
c = d.k
\end{array} \right.\\
 \Leftrightarrow \dfrac{{2a + 13b}}{{3a - 7b}} = \dfrac{{2.b.k + 13b}}{{3.b.k - 7.b}} = \dfrac{{2k + 13}}{{3k - 7}}\\
\dfrac{{2c + 13d}}{{3c - 7d}} = \dfrac{{2.d.k + 13d}}{{3.d.k - 7d}} = \dfrac{{2k + 13}}{{3k - 7}}\\
 \Leftrightarrow \dfrac{{2a + 13b}}{{3a - 7b}} = \dfrac{{2c + 13d}}{{3c - 7d}}\left( {tmdk} \right)\\
Vậy\,\dfrac{a}{b} = \dfrac{c}{d}
\end{array}$