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Question

Cho $a^2+b^2+c^2=m$

Tính GTBT theo m:

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

kudo shinichi · Accepted Answer

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

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

$A=\left[2.\left(a+b+c\right)-3c\right]^2+\left[2.\left(a+b+c\right)-3a\right]^2+\left[2.\left(a+b+c\right)-3b\right]^2$

Đặt $a+b+c=n$

$\Rightarrow A=\left(2n-3c\right)^2+\left(2n-3a\right)^2+\left(2n-3b\right)$

$A=4n^2-12cn+9c^2+4n^2-12an+9a^2+4n^2-12bn+9b^2$

$A=12n.\left(n-a-b-c\right)+9.\left(a^2+b^2+c^2\right)$

Ta có: $a^2+b^2+c^2=m$

$\Rightarrow A=12.\left(a+b+c-a-b-c\right)+9m$

$A=9m$

Vậy $A=9m$tại $a^2+b^2+c^2=m$

Tham khảo nhé~

Câu trả lời:

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

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

$A=\left[2.\left(a+b+c\right)-3c\right]^2+\left[2.\left(a+b+c\right)-3a\right]^2+\left[2.\left(a+b+c\right)-3b\right]^2$

Đặt $a+b+c=n$

$\Rightarrow A=\left(2n-3c\right)^2+\left(2n-3a\right)^2+\left(2n-3b\right)$

$A=4n^2-12cn+9c^2+4n^2-12an+9a^2+4n^2-12bn+9b^2$

$A=12n.\left(n-a-b-c\right)+9.\left(a^2+b^2+c^2\right)$

Ta có: $a^2+b^2+c^2=m$

$\Rightarrow A=12.\left(a+b+c-a-b-c\right)+9m$

$A=9m$

Vậy $A=9m$tại $a^2+b^2+c^2=m$

Tham khảo nhé~