Ta có: \(a+b+c=0\)
\(=>\left(a+b+c\right)^2=0\)
\(=>a^2+b^2+c^2+2ab+2bc+2ac=0\)
\(=>a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)
\(=>a^2+b^2+c^2=0\)
\(=>a^2+b^2+c^2=ab+bc+ac\)
\(=>2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ac\right)\)
\(=>\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)(nhân phân phối, đổi qua bên kia dấu bằng, tách thành hằng đẳng thức)
\(=>\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(=>\hept{\begin{cases}a-b=0\\b-c=0\\a-c=0\end{cases}}\)
\(=>a=b=c=0\)
***\(A=\left(a-1\right)^{22}+b^{12}+\left(c-1\right)^{2014}\)
\(A=\left(-1\right)^{22}+1+\left(-1\right)^{2014}\)
\(A=1+1+1\)
\(A=3\)
Ta có
a + b + c = 0
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow\)a2 + b2 + c2 = ab + bc + ca
Mà ta có a2 + b2 + c2 \(\ge\) ab + bc + ca
Dấu = xảy ra khi a = b = c = 0
\(\Rightarrow\)(a - 1)22 + b12 + (c - 1)2014 = 1 + 0 + 1 = 2
