Question

Answer 1

Xét pt :

\(x^2+2x-3=0\)

Theo định lí Viet ta có :

\(\left\{{}\begin{matrix}x_1+x_2=-2\\x_1.x_2=-3\end{matrix}\right.\)

a/ \(A=\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_1+x_2}{x_1.x_2}=\dfrac{2}{3}\)

b/ \(B=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1.x_2=2^2+2.3=10\)

c/ \(C=\dfrac{1}{x_1^2}+\dfrac{1}{x_2^2}=\dfrac{x_1^2+x_2^2}{x_1^2.x_2^2}=\dfrac{10}{9}\)

d/ \(D=x_1^3+x_2^3=\left(x_1+x_2\right)\left(x_1^2-x_1.x_2+x_2^2\right)=\left(x_1+x_2\right)\left[\left(x_1+x_2\right)^2-3x_1.x_2\right]=-26\)