Question

Giúp mình giải câu 2 với

Answer 1

Theo đ/l Viete, ta có:

\(\left\{{}\begin{matrix}S=x_1+x_2=\dfrac{-b}{a}=-\dfrac{3}{2}\\P=x_1x_2=\dfrac{c}{a}=-\dfrac{1}{2}\end{matrix}\right.\)

\(A=\dfrac{x_1}{x_2}\left(1-x_2\right)+\dfrac{x_2}{x_1}\left(1-x_1\right)=\dfrac{x_1}{x_2}-\dfrac{x_1x_2}{x_2}+\dfrac{x_2}{x_1}-\dfrac{x_1x_2}{x_1}=\dfrac{-\left(x_1^2+x_2^2\right)-2x_1x_2}{x_1x_2}=\dfrac{-S^2+2P-2P}{P}=\dfrac{-\left(-\dfrac{3}{2}\right)^2+2\left(-\dfrac{1}{2}\right)-2\left(-\dfrac{1}{2}\right)}{-\dfrac{1}{2}}=\dfrac{9}{2}\)