Theo Vi-et, ta có:
\(x_1+x_2=-\dfrac{b}{a}=-\dfrac{7}{5};x_1x_2=\dfrac{c}{a}=\dfrac{2}{5}\)
\(A=\dfrac{x_1}{5x_2^2+8x_2}+\dfrac{x_2}{5x_1^2+8x_1}\)
\(=\dfrac{x_1\left(5x_1^2+8x_1\right)+x_2\left(5x_2^2+8x_2\right)}{\left(5x_1^2+8x_1\right)\left(5x_2^2+8x_2\right)}\)
\(=\dfrac{5\left(x_1^3+x_2^3\right)+8\left(x_1^2+x_2^2\right)}{25\cdot\left(x_1x_2\right)^2+40\cdot x_1x_2^2+40\cdot x_1^2\cdot x_2+64x_1x_2}\)
\(=\dfrac{5\left[\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\right]+8\left[\left(x_1+x_2\right)^2-2x_1x_2\right]}{25\cdot\left(x_1x_2\right)^2+40\cdot x_1x_2\left(x_1+x_2\right)+64x_1x_2}\)
\(=\dfrac{5\cdot\left[\left(-1,4\right)^3-3\cdot\left(-1,4\right)\cdot0,4\right]+8\left[\left(-1,4\right)^2-2\cdot0,4\right]}{25\cdot\left(0,4\right)^2+40\cdot0,4\cdot\left(-1,4\right)+64\cdot0,4}\)
\(=\dfrac{11}{20}\)
