Đáp án:
\(Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = - \dfrac{2}{5}\\
{x_1}.{x_2} = \dfrac{{ - 16}}{5}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
Do:\Delta = 81 - 4.2.7 = 25\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = - \dfrac{9}{2}\\
{x_1}.{x_2} = \dfrac{7}{2}
\end{array} \right.\\
d)Do:\Delta ' = 9 - \dfrac{{168}}{{25}} = \dfrac{{57}}{{25}}\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{3}{{1,4}} = \dfrac{{15}}{7}\\
{x_1}.{x_2} = \dfrac{{1,2}}{{1,4}} = \dfrac{6}{7}
\end{array} \right.\\
f)Do:\Delta ' = 1 - 5.\left( { - 16} \right) = 81\\
\to \left[ \begin{array}{l}
x = \dfrac{{ - 1 + \sqrt {81} }}{5}\\
x = \dfrac{{ - 1 - \sqrt {81} }}{5}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{8}{5}\\
x = - 2
\end{array} \right.\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = - \dfrac{2}{5}\\
{x_1}.{x_2} = \dfrac{{ - 16}}{5}
\end{array} \right.
\end{array}\)
