Đáp án:
\(A = \frac{{3{{\left( {{x_1} + {x_2}} \right)}^2} - {x_1}{x_2}}}{{8{x_1}{x_2}.{x_2}}}\)
Giải thích các bước giải:
\(\begin{array}{l}
3{x^2} + 17x - 14 = 0\\
\to \left[ \begin{array}{l}
x = \frac{{ - 17 + \sqrt {457} }}{6}\\
x = \frac{{ - 17 - \sqrt {457} }}{6}
\end{array} \right.\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = - \frac{{17}}{3}\\
{x_1}{x_2} = - \frac{{14}}{3}
\end{array} \right.\\
A = \frac{{3{x_1}^2 + 3{x_2}^2 + 5{x_1}{x_2}}}{{4{x_1}{x_2}^2 + 4{x_1}{x_2}^2}}\\
= \frac{{3\left( {{x_1}^2 + {x_2}^2} \right) + 5{x_1}{x_2}}}{{8{x_1}{x_2}.{x_2}}}\\
= \frac{{3\left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2}} \right) + 5{x_1}{x_2}}}{{8{x_1}{x_2}.{x_2}}}\\
= \frac{{3{{\left( {{x_1} + {x_2}} \right)}^2} - 6{x_1}{x_2} + 5{x_1}{x_2}}}{{8{x_1}{x_2}.{x_2}}}\\
= \frac{{3{{\left( {{x_1} + {x_2}} \right)}^2} - {x_1}{x_2}}}{{8{x_1}{x_2}.{x_2}}}\\
\to \left[ \begin{array}{l}
A = \frac{{3{{\left( { - \frac{{17}}{3}} \right)}^2} + \frac{{14}}{3}}}{{8.\left( { - \frac{{14}}{3}} \right).\frac{{ - 17 + \sqrt {457} }}{6}}}\\
A = \frac{{3{{\left( { - \frac{{17}}{3}} \right)}^2} + \frac{{14}}{3}}}{{8.\left( { - \frac{{14}}{3}} \right).\frac{{ - 17 - \sqrt {457} }}{6}}}
\end{array} \right.\\
\to \left[ \begin{array}{l}
A = \frac{{101}}{{ - 27,2381407}}\\
A = \frac{{101}}{{238,7936963}}
\end{array} \right.
\end{array}\)
