Đáp án:
$\begin{array}{l}
\dfrac{a}{b} = \dfrac{c}{d} = k \Leftrightarrow \left\{ \begin{array}{l}
a = b.k\\
c = d.k
\end{array} \right.\\
a){\left( {\dfrac{{a - b}}{{c - d}}} \right)^4} = {\left( {\dfrac{{bk - b}}{{dk - d}}} \right)^4} = {\left( {\dfrac{b}{d}} \right)^4} = \dfrac{{{b^4}}}{{{d^4}}}\\
\dfrac{{{a^4} + {b^4}}}{{{c^4} + {d^4}}} = \dfrac{{{{\left( {bk} \right)}^4} + {b^4}}}{{{{\left( {dk} \right)}^4} + {d^4}}} = \dfrac{{{b^4}}}{{{d^4}}}\\
\Leftrightarrow {\left( {\dfrac{{a - b}}{{c - d}}} \right)^4} = \dfrac{{{a^4} + {b^4}}}{{{c^4} + {d^4}}}\\
b)\\
\dfrac{{5a + 3b}}{{5a - 3b}} = \dfrac{{5.bk + 3b}}{{5bk - 3b}} = \dfrac{{5k + 3}}{{5k - 3}}\\
\dfrac{{5c + 3d}}{{5c - 3d}} = \dfrac{{5.dk + 3d}}{{5dk - 3d}} = \dfrac{{\left( {5k + 3} \right)}}{{5k - 3}}\\
\Leftrightarrow \dfrac{{5a + 3b}}{{5a - 3b}} = \dfrac{{5c + 3d}}{{5c - 3d}}\\
c)\dfrac{{7{a^2} + 3ab}}{{11{a^2} - 8{b^2}}} = \dfrac{{7.{{\left( {bk} \right)}^2} + 3.bk.b}}{{11.{{\left( {bk} \right)}^2} - 8.{b^2}}}\\
= \dfrac{{{b^2}\left( {7{k^2} + 3k} \right)}}{{{b^2}\left( {11{k^2} - 8} \right)}}\\
= \dfrac{{7{k^2} + 3k}}{{11{k^2} - 8}}\\
\dfrac{{7{c^2} + 3cd}}{{11{c^2} - 8{d^2}}} = \dfrac{{7.{{\left( {dk} \right)}^2} + 3.dk.d}}{{11.{{\left( {dk} \right)}^2} - 8.{d^2}}}\\
= \dfrac{{{d^2}\left( {7{k^2} + 3k} \right)}}{{{d^2}\left( {11{k^2} - 8} \right)}}\\
= \dfrac{{7{k^2} + 3k}}{{11{k^2} - 8}}\\
\Leftrightarrow \dfrac{{7{a^2} + 3ab}}{{11{a^2} - 8{b^2}}} = \dfrac{{7{c^2} + 3cd}}{{11{c^2} - 8{d^2}}}
\end{array}$
