Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-\dfrac{4}{3}\\x_1x_2=\dfrac{c}{a}=-\dfrac{1}{3}\end{matrix}\right.\)
\(A=\dfrac{x_1^3\cdot x_2^2\left(1-3x_2^2\right)}{x_1^2+x_2^2}\)
\(=\dfrac{x_1^3\cdot x_2^2\cdot4x_2}{\left(x_1+x_2\right)^2-2x_1x_2}=\dfrac{4\cdot\left(x_1x_2\right)^3}{\left(x_1+x_2\right)^2-2x_1x_2}\)
\(=\dfrac{4\cdot\left(-\dfrac{1}{3}\right)^3}{\left(-\dfrac{4}{3}\right)^2-2\cdot\dfrac{-1}{3}}=\dfrac{-4}{27}:\left(\dfrac{16}{9}+\dfrac{2}{3}\right)\)
\(=-\dfrac{4}{27}:\dfrac{22}{9}=-\dfrac{4}{27}\cdot\dfrac{9}{22}=\dfrac{-2}{11}\cdot\dfrac{1}{3}=-\dfrac{2}{33}\)
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