Đáp án:
$\begin{array}{l}
\text{Đặt}:\dfrac{a}{b} = \dfrac{c}{d} = k \Rightarrow \left\{ \begin{array}{l}
a = b.k\\
c = d.k
\end{array} \right.\\
1)\\
\dfrac{{a + b}}{b} = \dfrac{{b.k + b}}{b} = \dfrac{{b.\left( {k + 1} \right)}}{b} = k + 1\\
\dfrac{{c + d}}{d} = \dfrac{{d.k + d}}{d} = \dfrac{{d\left( {k + 1} \right)}}{d} = k + 1\\
\Rightarrow \dfrac{{a + b}}{b} = \dfrac{{c + d}}{d}\left( { = k + 1} \right)\\
\text{Vậy}\,\dfrac{{a + b}}{b} = \dfrac{{c + d}}{d}\\
2)\dfrac{{2a + 3b}}{{2a - 3b}} = \dfrac{{2.b.k + 3b}}{{2.b.k - 3.b}}\\
= \dfrac{{b.\left( {2k + 3} \right)}}{{b.\left( {2k - 3} \right)}} = \dfrac{{2k + 3}}{{2k - 3}}\\
\dfrac{{2c + 3d}}{{2c - 3d}} = \dfrac{{2.d.k + 3.d}}{{2.d.k - 3d}}\\
= \dfrac{{d.\left( {2k + 3} \right)}}{{d.\left( {2k - 3} \right)}} = \dfrac{{2k + 3}}{{2k - 3}}\\
\Rightarrow \dfrac{{2a + 3b}}{{2a - 3b}} = \dfrac{{2c + 3d}}{{2c - 3d}}\left( { = \dfrac{{2k + 3}}{{2k - 3}}} \right)\\
\text{Vậy}\,\dfrac{{2a + 3b}}{{2a - 3b}} = \dfrac{{2c + 3d}}{{2c - 3d}}
\end{array}$
