Đáp án:
b) \(Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = - \dfrac{9}{2}\\
{x_1}.{x_2} = \dfrac{7}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Do:\Delta = 49 - 4.2.2 = 33\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = - \dfrac{{ - 7}}{2} = \dfrac{7}{2}\\
{x_1}.{x_2} = \dfrac{2}{2} = 1
\end{array} \right.\\
c)Do:\Delta ' = 4 - \left( {2 + \sqrt 2 } \right)\left( {2 - \sqrt 3 } \right) = 2\sqrt 3 - 2\sqrt 2 + \sqrt 6 \\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = - \dfrac{4}{{2 - \sqrt 3 }} = \dfrac{{ - 4\left( {2 + \sqrt 3 } \right)}}{{4 - 3}} = - 8 - 4\sqrt 3 \\
{x_1}.{x_2} = \dfrac{{2 + \sqrt 2 }}{{2 - \sqrt 3 }} = \dfrac{{\left( {2 + \sqrt 2 } \right)\left( {2 + \sqrt 3 } \right)}}{{4 - 3}} = 4 + 2\sqrt 3 + 2\sqrt 2 + \sqrt 6
\end{array} \right.
\end{array}\)
\(\begin{array}{l}
b)Do:\Delta = 81 - 4.2.7 = 25\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = - \dfrac{9}{2}\\
{x_1}.{x_2} = \dfrac{7}{2}
\end{array} \right.\\
d)Do:\Delta ' = 9 - \dfrac{{168}}{{25}} = \dfrac{{57}}{{25}}\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{3}{{1,4}} = \dfrac{{15}}{7}\\
{x_1}.{x_2} = \dfrac{{1,2}}{{1,4}} = \dfrac{6}{7}
\end{array} \right.
\end{array}\)
