Question

bình phương của \(|x1-x2|^{ }_{ }\)?

Answer 1

Ta có : \(\left(\left|x_1+x_2\right|\right)^2=\left(x_1+x_2\right)^2=x^2_1+2x_1x_2+x_2^2\)

Answer 2

\(\left(\left|x_1-x_2\right|\right)^2=x_1^2-2\left|x_1x_2\right|+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2-2\left|x_1x_2\right|\)