\(\text{Δ}=\left(-6\right)^2-4\cdot3\cdot2=36-24=12>0\)
=>(1) có hai nghiệm phân biệt
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{6}{3}=2\\x_1x_2=\dfrac{c}{a}=\dfrac{2}{3}\end{matrix}\right.\)
\(A=x_1^2+x_2^2-x_1x_2\)
\(=\left(x_1+x_2\right)^2-3x_1x_2\)
\(=2^2-3\cdot\dfrac{2}{3}=4-2=2\)