Câu hỏi của Nguyễn Minh Anh

\(A=\left(x_1+x_2\right)^2-2x_1x_2=\left(\dfrac{1}{3}\right)^2-2\cdot\dfrac{-1}{3}=\dfrac{1}{9}+\dfrac{2}{3}=\dfrac{7}{9}\simeq0.77\)Câu trả lời: \(A=\left(x_1+x_2\right)^2-2x_1x_2=\left(\dfrac{1}{3}\right)^2-2\cdot\dfrac{-1}{3}=\dfrac{1}{9}+\dfrac{2}{3}=\dfrac{7}{9}\simeq0.77\)