Đáp án:
$\begin{array}{l}
\dfrac{a}{b} = \dfrac{c}{d} = k \Leftrightarrow \left\{ \begin{array}{l}
a = b.k\\
c = d.k
\end{array} \right.\\
a)\dfrac{{a + b}}{b} = \dfrac{a}{b} + \dfrac{b}{b} = \dfrac{a}{b} + 1\\
\dfrac{{c + d}}{d} = \dfrac{c}{d} + \dfrac{d}{d} = \dfrac{c}{d} + 1\\
Do:\dfrac{a}{b} = \dfrac{c}{d}\\
\Leftrightarrow \dfrac{a}{b} + 1 = \dfrac{c}{d} + 1\\
\Leftrightarrow \dfrac{{a + b}}{b} = \dfrac{{c + d}}{d}\\
b)\dfrac{{a - b}}{b} = \dfrac{a}{b} - \dfrac{b}{b} = \dfrac{a}{b} - 1\\
\dfrac{{c - d}}{d} = \dfrac{c}{d} - 1 = \dfrac{a}{b} - 1\\
\Leftrightarrow \dfrac{{a - b}}{b} = \dfrac{{c - d}}{d}\\
c)\dfrac{{a + b}}{a} = \dfrac{a}{a} + \dfrac{b}{a} = 1 + \dfrac{b}{a}\\
\dfrac{{c + d}}{c} = \dfrac{c}{c} + \dfrac{d}{c} = 1 + \dfrac{d}{c}\\
Do:\dfrac{a}{b} = \dfrac{c}{d}\\
\Leftrightarrow \dfrac{b}{a} = \dfrac{d}{c}\\
\Leftrightarrow 1 + \dfrac{b}{a} = 1 + \dfrac{d}{c}\\
\Leftrightarrow \dfrac{{a + b}}{a} = \dfrac{{c + d}}{c}\\
d)\dfrac{a}{{a + b}} = \dfrac{{b.k}}{{b.k + b}} = \dfrac{{b.k}}{{b.\left( {k + 1} \right)}} = \dfrac{k}{{k + 1}}\\
\dfrac{c}{{c + d}} = \dfrac{{d.k}}{{d.k + d}} = \dfrac{{dk}}{{d\left( {k + 1} \right)}} = \dfrac{k}{{k + 1}}\\
\Leftrightarrow \dfrac{a}{{a + b}} = \dfrac{c}{{c + d}}
\end{array}$
