Question

Cho \(a^2+b^2+c^2=m\) . Tính giá trị của biểu thức sau theo m:

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

Answer 1

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

\(A=\left(2a+2b+2c-3x\right)^2+\left(2b+2c+2a-3a\right)^2+\left(2c+2a+2b-3b\right)^2\)

Đặt a + b + c = x thì:

\(A=\left(2x-3c\right)^2+\left(2x-3a\right)^2+\left(2x-3b\right)^2\)

\(=4x^2-12cx+9c^2+4x^2-12ax+9a^2+4x^2-12bx+9b^2\)

\(=12x^2-12x\left(a+b+c\right)+9\left(a^2+b^2+c^2\right)\)

\(12x^2-12x^2+9\left(a^2+b^2+c^2\right)=9\left(a^2+b^2+c^2\right)=9m\)

Answer 2

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

\(A=4a^2+4b^2+c^2+8ab-4bc-4ac+4b^2+4c^2+a^2+8ac-4ca-4ba+4c^2+4a^2+b^2+8ca-4ab-4cb\)

\(A=9a^2+9b^2+9c^2=9\left(a^2+b^2+c^2\right)=9m\)