Theo Vi-et, ta có:
\(x_1+x_2=-\dfrac{b}{a}=-\dfrac{2}{3};x_1x_2=\dfrac{c}{a}=-\dfrac{4}{3}\)
\(A=\dfrac{3x_1^2+5x_1x_2+3x_2^2}{4x_1^3\cdot x_2+4x_1\cdot x_2^3}\)
\(=\dfrac{3\left(x_1^2+x_2^2\right)+5\cdot\dfrac{-4}{3}}{4x_1x_2\left(x_1^2+x_2^2\right)}\)
\(=\dfrac{3\left[\left(x_1+x_2\right)^2-2x_1x_2\right]+\dfrac{-20}{3}}{4\cdot\dfrac{-4}{3}\cdot\left[\left(x_1+x_2\right)^2-2x_1x_2\right]}\)
\(=\dfrac{3\left[\left(-\dfrac{2}{3}\right)^2-2\cdot\dfrac{-4}{3}\right]+\dfrac{-20}{3}}{\dfrac{-16}{3}\left[\left(-\dfrac{2}{3}\right)^2-2\cdot\dfrac{-4}{3}\right]}=\dfrac{-9}{56}\)
Theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{2}{3}\\x_1x_2=-\dfrac{4}{3}\end{matrix}\right.\)
\(A=\dfrac{3\left(x_1+x_2\right)^2-x_1x_2}{4x_1x_2\left(x_1^2+x_2^2\right)}=\dfrac{3\left(x_1+x_2\right)^2-x_1x_2}{4x_1x_2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]}\)
\(=\dfrac{3\left(-\dfrac{2}{3}\right)^2-\left(-\dfrac{4}{3}\right)}{4.\left(-\dfrac{4}{3}\right).\left[\left(-\dfrac{2}{3}\right)^2-2.\left(-\dfrac{4}{3}\right)\right]}=-\dfrac{9}{56}\)
