Question

[Núi non tình yêu thuần khiết]

Cho \(a^2+b^2+c^2=m\)

Tính GTBT theo m:

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

Answer 1

[Akai Haruma]

Lời giải:

Đặt \(a+b+c=t\)

\(A=(2a+2b-c)^2+(2b+2c-a)^2+(2c+2a-b)^2\)

\(=(2a+2b+2c-3c)^2+(2b+2c+2a-3a)^2+(2c+2a+2b-3b)^2\)

\(=(2t-3c)^2+(2t-3a)^2+(2t-3b)^2\)

\(=4t^2+9c^2-12tc+4t^2+9a^2-12ta+4t^2+9b^2-12tb\)

\(=12t^2+9(a^2+b^2+c^2)-12t(a+b+c)\)

\(=12t^2+9m-12t^2=9m\)