Pt phải là: \(x^2+3\sqrt{3}x+1=0\)
\(\Delta=\left(3\sqrt{3}\right)^2-4.1=27-4=23>0\)
=> pt có 2 nghiệm phân biệt
Theo Vi-et:
\(\left\{{}\begin{matrix}x_1+x_2=-\frac{b}{a}=-3\sqrt{3}\\x_1x_2=\frac{c}{a}=1\end{matrix}\right.\)
\(A=\frac{3\left(x_1+x_2\right)^2-6x_1x_2+5x_1x_2}{4x_1x_2\left(x_1^2+x_2^2\right)}\)
\(A=\frac{3\left(x_1+x_2\right)^2-x_1x_2}{4x_1x_2\left(x_1+x_2\right)^2-8\left(x_1x_2\right)^2}\)
\(A=\frac{3.\left(-3\sqrt{3}\right)^2-1}{4.1.\left(-3\sqrt{3}\right)^2-8.1}=\frac{4}{5}\)