Đáp án:
b. \(\frac{{661}}{{10}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.Vi - et:\\
\left\{ \begin{array}{l}
{x_1} + {x_2} = \frac{{13}}{2}\\
{x_1}{x_2} = \frac{{20}}{2} = 10
\end{array} \right.\\
b.3\left( {{x_1}^2 + {x_2}^2} \right) - \left( {\frac{{{x_2} + {x_1}}}{{{x_1}{x_2}}}} \right)\\
= 3\left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2}} \right) - \left( {\frac{{{x_2} + {x_1}}}{{{x_1}{x_2}}}} \right)\\
= 3{\left( {{x_2} + {x_1}} \right)^2} - 6{x_1}{x_2} - \left( {\frac{{{x_2} + {x_1}}}{{{x_1}{x_2}}}} \right)\\
= 3{\left( {\frac{{13}}{2}} \right)^2} - 6.10 - \frac{{13}}{2}:10\\
= \frac{{507}}{4} - 60 - \frac{{13}}{{20}} = \frac{{661}}{{10}}
\end{array}\)
