Đáp án:
$\begin{array}{l}
2)Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{{ - b}}{a}\\
{x_1}{x_2} = \dfrac{c}{a}
\end{array} \right.\\
a)2{x^2} - 7x + 2 = 0\\
\Rightarrow \left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{7}{2}\\
{x_1}{x_2} = 1
\end{array} \right.\\
b)2{x^2} + 9x + 7 = 0\\
\Rightarrow \left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{{ - 9}}{2}\\
{x_1}{x_2} = \dfrac{7}{2}
\end{array} \right.\\
c)\left( {2 - \sqrt 3 } \right){x^2} + 4x + 2 + \sqrt 2 = 0\\
\Rightarrow \left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{4}{{\sqrt 3 - 2}} = \dfrac{{4\left( {\sqrt 3 + 2} \right)}}{{ - 1}} = - 4\sqrt 3 - 8\\
{x_1}{x_2} = \dfrac{{2 + \sqrt 2 }}{{2 - \sqrt 3 }} = \dfrac{{\left( {2 + \sqrt 2 } \right)\left( {2 + \sqrt 3 } \right)}}{1} = 4 + 2\sqrt 3 + 2\sqrt 2 + \sqrt 6
\end{array} \right.\\
d)1,4{x^2} - 3x + 1,2 = 0\\
\Rightarrow \left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{3}{{1,4}} = \dfrac{{15}}{7}\\
{x_1}{x_2} = \dfrac{{1,2}}{{1,4}} = \dfrac{6}{7}
\end{array} \right.
\end{array}$
