Theo Vi-ét: `x_1 + x_2 = 11/3 ; x_1 x_2 = -5`
\(A=\dfrac{3x_1}{x_2}+\dfrac{3x_2}{x_1}\\ =\dfrac{3x_1^2+3x_2^2}{x_1x_2}\\ =\dfrac{3\left(x_1^2+x_2^2\right)}{x_1x_2}\\ =\dfrac{3\left[\left(x_1+x_2\right)^2-2x_1x\right]_2}{x_1x_2}\\ =\dfrac{3\left(x_1+x_2\right)^2-6x_1x_2}{x_1x_2}\\ =\dfrac{3.\left(\dfrac{11}{3}\right)^2-6\left(-5\right)}{-5}\\ =-\dfrac{211}{15}\)