Question

Cho \(a^2+b^2+c^2=m\) . Tính giá trị 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=(2a+2b-c)^2+(2b+2c-a)^2+(2c+2a-b)^2`

`=[(2a)^2+(2b)^2+c^2+2*2a*2b+2*2a*(-c)+2*2b*(-c)]+[(2b)^2+(2c)^2+a^2+2*2b*2c+2*2b*(-a)+2*2c*(-a)]+[(2a)^2+(2c)^2+b^2+2*2a*2c+2*2a*(-b)+2*2c*(-b)]`

`=(4a^2+4b^2+c^2+8ab-4ac-4bc)+(4b^2+4c^2+a^2+8bc-4ac-4ab)+(4a^2+4c^2+b^2+8ac-4bc-4ba)`

`=(4a^2+a^2+4a^2)+(4b^2+b^2+4b^2)+(4c^2+c^2+4c^2)+(8ab-4ab-4ab)+(8ac-4ac-4ac)+(8bc-4bc-4bc)`

`=9a^2+9b^2+9c^2`

`=9(a^2+b^2+c^2)`

`=9m`

Vậy: `A=9m`