Đáp án:
$\begin{array}{l}
\dfrac{a}{b} = \dfrac{c}{d} = k \Leftrightarrow \left\{ \begin{array}{l}
a = b.k\\
c = d.k
\end{array} \right.\\
3)\dfrac{{{a^2} + {b^2}}}{{{a^2} - {b^2}}} = \dfrac{{{{\left( {bk} \right)}^2} + {b^2}}}{{{{\left( {bk} \right)}^2} - {b^2}}}\\
= \dfrac{{{b^2}\left( {{k^2} + 1} \right)}}{{{b^2}\left( {{k^2} - 1} \right)}} = \dfrac{{{k^2} + 1}}{{{k^2} - 1}}\\
\dfrac{{{c^2} + {d^2}}}{{{c^2} - {d^2}}} = \dfrac{{{{\left( {dk} \right)}^2} + {d^2}}}{{{{\left( {dk} \right)}^2} - {d^2}}} = \dfrac{{{d^2}\left( {{k^2} + 1} \right)}}{{{d^2}\left( {{k^2} - 1} \right)}} = \dfrac{{{k^2} + 1}}{{{k^2} - 1}}\\
\Leftrightarrow \dfrac{{{a^2} + {b^2}}}{{{a^2} - {b^2}}} = \dfrac{{{c^2} + {d^2}}}{{{c^2} - {d^2}}}\\
4)\\
+ )\dfrac{{{{\left( {a - b} \right)}^3}}}{{{{\left( {c - d} \right)}^3}}} = \dfrac{{{{\left( {bk - b} \right)}^3}}}{{{{\left( {dk - d} \right)}^3}}}\\
= \dfrac{{{b^3}{{\left( {k - 1} \right)}^3}}}{{{d^3}{{\left( {k - 1} \right)}^3}}} = \dfrac{{{b^3}}}{{{d^3}}}\\
+ )\dfrac{{3{a^3} + 2{b^3}}}{{3{c^3} + 2{d^3}}} = \dfrac{{3{{\left( {bk} \right)}^3} + 2{b^3}}}{{3{{\left( {dk} \right)}^3} + 2{d^3}}}\\
= \dfrac{{{b^3}\left( {3{k^3} + 2} \right)}}{{{d^3}\left( {3{k^3} + 2} \right)}} = \dfrac{{{b^3}}}{{{d^3}}}\\
\Leftrightarrow \dfrac{{{{\left( {a - b} \right)}^3}}}{{{{\left( {c - d} \right)}^3}}} = \dfrac{{3{a^3} + 2{b^3}}}{{3{c^3} + 2{d^3}}}
\end{array}$
