Đáp án:
\[\frac{{{{\left( {a + c} \right)}^2}}}{{{a^2} - {c^2}}} = \frac{{{{\left( {b + d} \right)}^2}}}{{{b^2} - {d^2}}} \Leftrightarrow \frac{a}{b} = \frac{c}{d}\]
Giải thích các bước giải:
\[\begin{array}{l}
\dfrac{{{{\left( {a + c} \right)}^2}}}{{{a^2} - {c^2}}} = \dfrac{{{{\left( {a + c} \right)}^2}}}{{\left( {a + c} \right)\left( {a - c} \right)}} = \dfrac{{a + c}}{{a - c}}\\
\dfrac{{{{\left( {b + d} \right)}^2}}}{{{b^2} - {d^2}}} = \dfrac{{{{\left( {b + d} \right)}^2}}}{{\left( {b - d} \right)\left( {b + d} \right)}} = \dfrac{{b + d}}{{b - d}}\\
\Rightarrow \dfrac{{{{\left( {a + c} \right)}^2}}}{{{a^2} - {c^2}}} = \dfrac{{{{\left( {b + d} \right)}^2}}}{{{b^2} - {d^2}}}\\
\Leftrightarrow \dfrac{{a + c}}{{a - c}} = \dfrac{{b + d}}{{b - d}}\\
\Leftrightarrow \left( {a + c} \right)\left( {b - d} \right) = \left( {a - c} \right)\left( {b + d} \right)\\
\Leftrightarrow ab - ad + bc - cd = ab + ad - bc - cd\\
\Leftrightarrow - 2ad = - 2bc\\
\Leftrightarrow \dfrac{a}{b} = \dfrac{c}{d}
\end{array}\]
