Giải thích các bước giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\begin{array}{l}
a,\\
\dfrac{a}{b} = \dfrac{c}{d} \Leftrightarrow \dfrac{a}{c} = \dfrac{b}{d} \Leftrightarrow \dfrac{{3a}}{{3c}} = \dfrac{{2b}}{{2d}}\\
\dfrac{{3a}}{{3c}} = \dfrac{{2b}}{{2d}} = \dfrac{{3a + 2b}}{{3c + 2d}} = \dfrac{{3a - 2b}}{{3c - 2d}}\\
\dfrac{{3a + 2b}}{{3c + 2d}} = \dfrac{{3a - 2b}}{{3c - 2d}} \Leftrightarrow \dfrac{{3a + 2b}}{{3a - 2b}} = \dfrac{{3c + 2d}}{{3c - 2d}}\\
b,\\
\dfrac{a}{b} = \dfrac{c}{d} \Leftrightarrow \dfrac{a}{c} = \dfrac{b}{d} = \dfrac{{a + b}}{{c + d}}\\
\Rightarrow {\left( {\dfrac{a}{c}} \right)^2} = {\left( {\dfrac{b}{d}} \right)^2} = {\left( {\dfrac{{a + b}}{{c + d}}} \right)^2} = \dfrac{{{{\left( {a + b} \right)}^2}}}{{{{\left( {c + d} \right)}^2}}}\\
{\left( {\dfrac{a}{c}} \right)^2} = \dfrac{a}{c}.\dfrac{a}{c} = \dfrac{a}{c}.\dfrac{b}{d} = \dfrac{{ab}}{{cd}}\\
\Rightarrow \dfrac{{ab}}{{cd}} = {\left( {\dfrac{a}{c}} \right)^2} = \dfrac{{{{\left( {a + b} \right)}^2}}}{{{{\left( {c + d} \right)}^2}}}
\end{array}\)
