\(\left\{{}\begin{matrix}x_1+x_2=3\\x_1x_2=2\end{matrix}\right.\)
\(A=\dfrac{2x_1^2+2x_2^2}{x_1x_2}+\dfrac{3}{4}=\dfrac{2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]}{2}+\dfrac{3}{4}\)
\(=\dfrac{2\cdot\left(3^2-2\cdot2\right)}{2}+\dfrac{3}{4}=9-4+0.75=5.75=\dfrac{23}{4}\)
Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=3\\x_1x_2=2\end{matrix}\right.\)
\(A=\dfrac{2x_1^2+2x_2^2}{x_1x_2}+\dfrac{3}{4}=\dfrac{2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]}{x_1x_2}+\dfrac{3}{4}\)
Thay vào ta được \(\dfrac{2\left(9-4\right)}{2}+\dfrac{3}{4}=5+\dfrac{3}{4}=\dfrac{23}{4}\)