Đáp án:
\(\dfrac{7}{{40}}\)
Giải thích các bước giải:
\(\begin{array}{l}
B2:\\
\Delta ' = 4.3 - 8 = 4\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = 4\sqrt 3 \\
{x_1}{x_2} = 8
\end{array} \right.\\
M = \dfrac{{6\left( {{x_1}^2 + {x_2}^2} \right) + 10{x_1}{x_2}}}{{5{x_1}{x_2}\left( {{x_1}^2 + {x_2}^2} \right)}}\\
= \dfrac{{6\left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2}} \right) + 10{x_1}{x_2}}}{{5{x_1}{x_2}\left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2}} \right)}}\\
= \dfrac{{6{{\left( {{x_1} + {x_2}} \right)}^2} - 2{x_1}{x_2}}}{{5{x_1}{x_2}\left( {{{\left( {{x_1} + {x_2}} \right)}^2} - 2{x_1}{x_2}} \right)}}\\
= \dfrac{{6.{{\left( {4\sqrt 3 } \right)}^2} - 2.8}}{{5.8\left( {{{\left( {4\sqrt 3 } \right)}^2} - 2.8} \right)}} = \dfrac{7}{{40}}
\end{array}\)
