Question

`2x^2 -5x-1=0` có 2 nghiệm `x_1 , x_2`. Tính \(\dfrac{x_1}{x_1-1}+\dfrac{x_2}{x_2-1}-2022\)

Answer 1

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{5}{2}\\x_1x_2=\dfrac{c}{a}=-\dfrac{1}{2}\end{matrix}\right.\)

\(\dfrac{x_1}{x_1-1}+\dfrac{x_2}{x_2-1}-2022\)

\(=\dfrac{x_1x_2-x_1+x_2x_1-x_2}{\left(x_1-1\right)\left(x_2-1\right)}-2022\)

\(=\dfrac{2\cdot x_1x_2-\left(x_1+x_2\right)}{x_1x_2-\left(x_1+x_2\right)+1}-2022\)

\(=\dfrac{2\cdot\dfrac{-1}{2}-\dfrac{5}{2}}{-\dfrac{1}{2}-\dfrac{-5}{2}+1}-2022\)

\(=\dfrac{-\dfrac{7}{2}}{-\dfrac{1}{2}+\dfrac{5}{2}+1}-2022=\dfrac{-7}{6}-2022=-\dfrac{12139}{6}\)