\(2x^2-5x-1=0\)
Theo Vi-ét:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{- \left(-5\right)}{2}=\dfrac{5}{2}\\x_1x_2=\dfrac{c}{a}=-\dfrac{1}{2}\end{matrix}\right.\)
Ta có : \(A=\dfrac{x_1}{x_1-1}+\dfrac{x_2}{x_2-1}-2022\)
\(=\dfrac{x_1\left(x_2-1\right)+x_2\left(x_1-1\right)}{\left(x_1-1\right)\left(x_2-1\right)}-2022\)
\(=\dfrac{x_1x_2-x_1+x_1x_2-x_2}{x_1x_2-x_1-x_2+1}-2022\)
\(=\dfrac{2x_1x_2-\left(x_1+x_2\right)}{x_1x_2-\left(x_1+x_2\right)+1}-2022\)
\(=\dfrac{2.\left(-\dfrac{1}{2}\right)-\dfrac{5}{2}}{-\dfrac{1}{2}-\dfrac{5}{2}+1}-2022\)
\(=-\dfrac{8081}{4}\)
Vậy \(A=-\dfrac{8081}{4}\)
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