Đáp án:
$\begin{array}{l}
\left\{ \begin{array}{l}
a.b = - 2\\
a + b = 2
\end{array} \right.\\
\Rightarrow a = 2 - b\\
\Rightarrow \left( {2 - b} \right).b = - 2\\
\Rightarrow 2b - {b^2} = - 2\\
\Rightarrow {b^2} - 2b - 2 = 0\\
\Rightarrow {b^2} - 2b + 1 = 3\\
\Rightarrow {\left( {b - 1} \right)^2} = 3\\
\Rightarrow \left[ \begin{array}{l}
b - 1 = \sqrt 3 \\
b - 1 = - \sqrt 3
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
b = 1 + \sqrt 3 \\
b = 1 - \sqrt 3
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
a = 2 - b = 1 - \sqrt 3 \\
a = 2 - b = 2 - \left( {1 - \sqrt 3 } \right) = 1 + \sqrt 3
\end{array} \right.\\
Vậy\,\left( {a;b} \right) = \left( {1 - \sqrt 3 ;1 + \sqrt 3 } \right)/\left( {1 + \sqrt 3 ;1 - \sqrt 3 } \right)
\end{array}$
