Theo Vi-et, ta có: \(\begin{cases}x_1+x_2=-\frac{b}{a}=\frac{-\left(-5\right)}{-2}=\frac{5}{-2}=-\frac52\\ x_1x_2=\frac{c}{a}=\frac{4}{-2}=-2\end{cases}\)
\(A=\left(x_1-x_2\right)^2-\frac{5}{2x_1}+\frac{x_1}{x_2}\)
\(=\left(x_1+x_2\right)^2-4x_1x_2+\frac{x_1+x_2}{x_1}+\frac{x_1}{x_2}\)
\(=\left(x_1+x_2\right)^2-4x_1x_2+\frac{x_1x_2+x_2^2+x_1^2}{x_1x_2}\)
\(=\left(x_1+x_2\right)^2-4x_1x_2+\frac{\left(x_1+x_2\right)^2-x_1x_2}{x_1x_2}\)
\(=\left(-2,5\right)^2-4\cdot\left(-2\right)+\frac{\left(-2,5\right)^2-\left(-2\right)}{-2}=6,25+8+\frac{6,25+2}{-2}=14,25-\frac{8.25}{2}=10,125\)
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