Đáp án:
$\begin{array}{l}
1)\\
A = {9^{n + 2}} + {3^{n + 2}} - {9^n} + {3^n}\\
= {9^n}{.9^2} - {9^n} + {3^n}{.3^2} + {3^n}\\
= {9^n}\left( {81 - 1} \right) + {3^n}.\left( {9 + 1} \right)\\
= {9^n}.80 + {3^n}.10\\
= \left( {{9^n}.8 + {3^n}} \right).10 \vdots 10\\
Vậy\,A \vdots 10\\
2)\\
Giả\,sử:\frac{a}{b} = \frac{c}{d} = k \Rightarrow \left\{ \begin{array}{l}
a = b.k\\
c = d.k
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {\frac{{a + b}}{{c + d}}} \right)^2} = {\left( {\frac{{b.k + b}}{{d.k + d}}} \right)^2} = \frac{{{b^2}.{{\left( {k + 1} \right)}^2}}}{{{d^2}{{\left( {k + 1} \right)}^2}}} = \frac{{{b^2}}}{{{d^2}}}\\
\frac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}} = \frac{{{b^2}.{k^2} + {b^2}}}{{{d^2}.{k^2} + {d^2}}} = \frac{{{b^2}\left( {{k^2} + 1} \right)}}{{{d^2}\left( {{k^2} + 1} \right)}} = \frac{{{b^2}}}{{{d^2}}}
\end{array} \right.\\
\Rightarrow {\left( {\frac{{a + b}}{{c + d}}} \right)^2} = \frac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}\left( { = \frac{{{b^2}}}{{{d^2}}}} \right)
\end{array}$
