Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\\x_1x_2=\dfrac{c}{a}=-\dfrac{1}{2}\end{matrix}\right.\)
\(T=\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}+\dfrac{5}{2}\left(x_1-x_2\right)^2\)
\(=\dfrac{x_1^2+x_2^2}{x_1x_2}+\dfrac{5}{2}\left[\left(x_1+x_2\right)^2-4x_1x_2\right]\)
\(=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1x_2}+\dfrac{5}{2}\cdot\left[\left(x_1+x_2\right)^2-4x_1x_2\right]\)
\(=\dfrac{2^2-2\cdot\dfrac{-1}{2}}{-\dfrac{1}{2}}+\dfrac{5}{2}\left[2^2-4\cdot\dfrac{-1}{2}\right]\)
\(=\dfrac{4+1}{-\dfrac{1}{2}}+\dfrac{5}{2}\cdot\left(4+2\right)\)
\(=-10+\dfrac{5}{2}\cdot6=-10+15=5\)
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