Đáp án:
$\begin{array}{l}
{m^2} = p.q\\
\Leftrightarrow \dfrac{m}{p} = \dfrac{q}{m} = k \Leftrightarrow \left\{ \begin{array}{l}
m = pk\\
q = mk = pk.k = p{k^2}
\end{array} \right.\\
+ \dfrac{{{m^2} + {q^2}}}{{{p^2} + {m^2}}} = \dfrac{{{m^2} + {m^2}{k^2}}}{{{p^2} + {p^2}{k^2}}} = \dfrac{{{m^2}}}{{{p^2}}} = {k^2}\\
+ \dfrac{{{m^2} - {q^2}}}{{{p^2} - {m^2}}} = \dfrac{{{m^2} - {m^2}{k^2}}}{{{p^2} - {p^2}{k^2}}} = \dfrac{{{m^2}}}{{{p^2}}} = {k^2}\\
+ \dfrac{{{{\left( {m - q} \right)}^2}}}{{{{\left( {p - m} \right)}^2}}} = \dfrac{{{{\left( {m - mk} \right)}^2}}}{{{{\left( {p - pk} \right)}^2}}} = \dfrac{{{m^2}{{\left( {1 - k} \right)}^2}}}{{{p^2}{{\left( {1 - k} \right)}^2}}} = {k^2}\\
+ \dfrac{{{{\left( {m + q} \right)}^2}}}{{{{\left( {p + m} \right)}^2}}} = \dfrac{{{{\left( {m + mk} \right)}^2}}}{{{{\left( {p + pk} \right)}^2}}} = \dfrac{{{m^2}{{\left( {1 + k} \right)}^2}}}{{{p^2}{{\left( {1 + k} \right)}^2}}} = {k^2}\\
+ \dfrac{q}{p} = \dfrac{{p{k^2}}}{p} = {k^2}\\
Vậy\,\dfrac{{{m^2} + {q^2}}}{{{p^2} + {m^2}}} = \dfrac{{{m^2} - {q^2}}}{{{p^2} - {m^2}}} = \dfrac{{{{\left( {m - q} \right)}^2}}}{{{{\left( {p - m} \right)}^2}}} = \dfrac{{{{\left( {m + q} \right)}^2}}}{{{{\left( {p + m} \right)}^2}}} = \dfrac{q}{p}
\end{array}$
