Theo Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{1}{2}\\x_1x_2=-2\end{matrix}\right.\) \(A=3-x_1^2-x_2^2\\ =3-\left(x_1^2+x_2^2\right)\\ =3-\left[\left(x_1+x_2\right)^2-2x_1x_2\right]\\ =3-\left[\left(-\dfrac{1}{2}\right)^2-2.\left(-2\right)\right]\\ =3-\left(\dfrac{1}{4}+4\right)\\ =3-\dfrac{17}{4}\\ =-\dfrac{5}{4}\) \(B=\left(x_1-x_2\right)^2\\ =x_1^2+x_2^2-2x_1x_2\\ =\left(x_1+x_2\right)^2-4x_1x_2\\ =\left(\dfrac{1}{2}\right)^2-4.\left(-2\right)\\ =\dfrac{1}{4}+8\\ =\dfrac{33}{4}\) \(D=\left(1+x_1\right)\left(2-x_1\right)+\left(1+x_2\right)\left(2-x_2\right)\\ =2+x_1-x_1^2+2+x_2-x_2^2\\ =4+\left(x_1+x_2\right)-\left(x_1^2+x_2^2\right)\\ =4+\dfrac{1}{2}-\left[\left(x_1+x_2\right)^2-2x_1x_2\right]\\ =\dfrac{9}{2}-\left[\left(\dfrac{1}{2}\right)^2-2.\left(-2\right)\right]\\ =\dfrac{9}{2}-\dfrac{17}{4}\\ =\dfrac{1}{4}\) Câu trả lời: Theo Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{1}{2}\\x_1x_2=-2\end{matrix}\right.\) \(A=3-x_1^2-x_2^2\\ =3-\left(x_1^2+x_2^2\right)\\ =3-\left[\left(x_1+x_2\right)^2-2x_1x_2\right]\\ =3-\left[\left(-\dfrac{1}{2}\right)^2-2.\left(-2\right)\right]\\ =3-\left(\dfrac{1}{4}+4\right)\\ =3-\dfrac{17}{4}\\ =-\dfrac{5}{4}\) \(B=\left(x_1-x_2\right)^2\\ =x_1^2+x_2^2-2x_1x_2\\ =\left(x_1+x_2\right)^2-4x_1x_2\\ =\left(\dfrac{1}{2}\right)^2-4.\left(-2\right)\\ =\dfrac{1}{4}+8\\ =\dfrac{33}{4}\) \(D=\left(1+x_1\right)\left(2-x_1\right)+\left(1+x_2\right)\left(2-x_2\right)\\ =2+x_1-x_1^2+2+x_2-x_2^2\\ =4+\left(x_1+x_2\right)-\left(x_1^2+x_2^2\right)\\ =4+\dfrac{1}{2}-\left[\left(x_1+x_2\right)^2-2x_1x_2\right]\\ =\dfrac{9}{2}-\left[\left(\dfrac{1}{2}\right)^2-2.\left(-2\right)\right]\\ =\dfrac{9}{2}-\dfrac{17}{4}\\ =\dfrac{1}{4}\)

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huy78

6 tháng 3 2022

giup minh vs

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ILoveMath

6 tháng 3 2022

Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{1}{2}\\x_1x_2=-2\end{matrix}\right.\)

\(A=3-x_1^2-x_2^2\\ =3-\left(x_1^2+x_2^2\right)\\ =3-\left[\left(x_1+x_2\right)^2-2x_1x_2\right]\\ =3-\left[\left(-\dfrac{1}{2}\right)^2-2.\left(-2\right)\right]\\ =3-\left(\dfrac{1}{4}+4\right)\\ =3-\dfrac{17}{4}\\ =-\dfrac{5}{4}\)

\(B=\left(x_1-x_2\right)^2\\ =x_1^2+x_2^2-2x_1x_2\\ =\left(x_1+x_2\right)^2-4x_1x_2\\ =\left(\dfrac{1}{2}\right)^2-4.\left(-2\right)\\ =\dfrac{1}{4}+8\\ =\dfrac{33}{4}\)

\(D=\left(1+x_1\right)\left(2-x_1\right)+\left(1+x_2\right)\left(2-x_2\right)\\ =2+x_1-x_1^2+2+x_2-x_2^2\\ =4+\left(x_1+x_2\right)-\left(x_1^2+x_2^2\right)\\ =4+\dfrac{1}{2}-\left[\left(x_1+x_2\right)^2-2x_1x_2\right]\\ =\dfrac{9}{2}-\left[\left(\dfrac{1}{2}\right)^2-2.\left(-2\right)\right]\\ =\dfrac{9}{2}-\dfrac{17}{4}\\ =\dfrac{1}{4}\)

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