Question

H=\(\left(a+b+c\right)^2\)+\(\left(a-b-c\right)^2\)+\(\left(b-c-a\right)^2\)+\(\left(c-a-b\right)^2\)

 

Answer 1

Xét ta có:

+) \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ac+2ab+2bc\)

+) \(\left(a-b-c\right)^2=a^2+b^2+c^2-2ac-2ab+2bc\)

+) \(\left(b-c-a\right)^2=a^2+b^2+c^2-2ab-2bc+2ac\)

+) \(\left(c-a-b\right)^2=a^2+b^2+c^2-2ac-2bc+2ab\)

Nên: \(H=\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2\)

\(H=a^2+b^2+c^2+2ac+2bc+2bc+a^2+b^2+c^2-2ac-2ab+2bc+a^2+b^2+c^2-2ab-2bc+2ac+a^2+b^2+c^2-2ac-2bc+2ab\)

\(H=\left(a^2+a^2+a^2\right)+\left(b^2+b^2+b^2\right)+\left(c^2+c^2+c^2\right)+\left(2ac-2ac+2ac-2ac\right)+\left(2ab-2ab-2ab+2ab\right)+\left(2bc+2bc-2bc-2bc\right)\)

\(H=3a^2+3b^2+3c^2+0+0+0\)

\(H=3\left(a^2+b^2+c^2\right)\)