Đáp án:
$\begin{array}{l}
\dfrac{a}{b} = \dfrac{c}{d} = k\\
\Rightarrow \left\{ \begin{array}{l}
a = b.k\\
c = d.k
\end{array} \right.\\
a)\dfrac{a}{{a + c}} = \dfrac{{b.k}}{{b.k + d.k}}\\
= \dfrac{{b.k}}{{\left( {b + d} \right).k}} = \dfrac{b}{{b + d}}\\
Vay\,\dfrac{a}{{a + c}} = \dfrac{b}{{b + d}}\\
b)\\
+ )\dfrac{{{a^2} + {c^2}}}{{{b^2} + {d^2}}} = \dfrac{{{b^2}{k^2} + {d^2}{k^2}}}{{{b^2} + {d^2}}}\\
= \dfrac{{\left( {{b^2} + {d^2}} \right).{k^2}}}{{{b^2} + {d^2}}} = {k^2}\\
+ )\dfrac{{ac}}{{bd}} = \dfrac{{b.k.d.k}}{{b.d}} = \dfrac{{bd.{k^2}}}{{bd}} = {k^2}\\
Vay\,\dfrac{{{a^2} + {c^2}}}{{{b^2} + {d^2}}} = \dfrac{{ac}}{{bd}}
\end{array}$
