Ta có: \(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\Leftrightarrow a=b=c\)
mà a+b+c=6
nên \(a=b=c=\frac{6}{3}=2\)
Vậy: \(A=\left(1-a\right)^{2017}+\left(b-1\right)^{2017}+\left(c-2\right)^{2017}\)
\(=\left(1-2\right)^{2017}+\left(2-1\right)^{2017}+\left(2-2\right)^{2017}\)
\(=-1^{2017}+1^{2017}=0\)
Đúng 0
Bình luận (0)
