Đáp án:
$\begin{array}{l}
a)Ta có:\,\dfrac{a}{b} = \dfrac{c}{d}\\
\Leftrightarrow \dfrac{a}{c} = \dfrac{b}{d} = \dfrac{{5a}}{{5c}} = \dfrac{{3a}}{{3c}} = \dfrac{{3b}}{{3d}} = \dfrac{{7b}}{{7d}}\\
= \dfrac{{5a + 3b}}{{5c + 3d}} = \dfrac{{3a - 7b}}{{3c - 7d}}\\
\Leftrightarrow \dfrac{{5a + 3b}}{{5c + 3d}} = \dfrac{{3a - 7b}}{{3c - 7d}}\\
\Leftrightarrow \dfrac{{5a + 7b}}{{3a - 7b}} = \dfrac{{5c + 3d}}{{3c - 7d}}\\
b)Ta có:\,\dfrac{a}{b} = \dfrac{c}{d}\\
\Leftrightarrow \dfrac{a}{c} = \dfrac{b}{d} = \dfrac{{a + b}}{{c + d}}\\
\Leftrightarrow \dfrac{{{a^2}}}{{{c^2}}} = \dfrac{{{b^2}}}{{{d^2}}} = {\left( {\dfrac{{a + b}}{{c + d}}} \right)^2} = \dfrac{{{a^2} + {b^2}}}{{{c^2} + {b^2}}}\\
\Leftrightarrow \dfrac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}} = {\left( {\dfrac{{a + b}}{{c + d}}} \right)^2}
\end{array}$
