Đáp án:
$\begin{array}{l}
a)A = \dfrac{{3 + \sqrt 5 }}{{3 - \sqrt 5 }} + \dfrac{{3 - \sqrt 5 }}{{3 + \sqrt 5 }}\\
= \dfrac{{{{\left( {3 + \sqrt 5 } \right)}^2} + {{\left( {3 - \sqrt 5 } \right)}^2}}}{{\left( {3 - \sqrt 5 } \right)\left( {3 + \sqrt 5 } \right)}}\\
= \dfrac{{9 + 6\sqrt 5 + 5 + 9 - 6\sqrt 5 + 5}}{{9 - 5}}\\
= \dfrac{{28}}{4}\\
= 7\\
2)a)m = 1\\
\Rightarrow \left\{ \begin{array}{l}
2x + y = 5.1 - 1\\
x - 2y = 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
2x + y = 4\\
x - 2y = 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
2x + y = 4\\
2x - 4y = 4
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
5y = 0\\
x = 2 + 2y
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
y = 0\\
x = 2
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left( {2;0} \right)\,khi\,m = - 1\\
b)\left\{ \begin{array}{l}
2x + y = 5m - 1\\
x - 2y = 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
2x + y = 5m - 1\\
2x - 4y = 4
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
5y = 5m - 1 - 4\\
x = 2y + 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
y = m - 1\\
x = 2\left( {m - 1} \right) + 2 = 2m
\end{array} \right.\\
Khi:{x^2} - 2{y^2} = 1\\
\Rightarrow {\left( {2m} \right)^2} - 2.{\left( {m - 1} \right)^2} = 1\\
\Rightarrow 4{m^2} - 2\left( {{m^2} - 2m + 1} \right) = 1\\
\Rightarrow 4{m^2} - 2{m^2} + 4m - 2 - 1 = 0\\
\Rightarrow 2{m^2} + 4m - 3 = 0\\
\Rightarrow m = \dfrac{{ - 2 \pm \sqrt {10} }}{2}\\
Vậy\,m = \dfrac{{ - 2 \pm \sqrt {10} }}{2}
\end{array}$
