Question

(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca

(a+b−c)2=a^2+b^2+c^2+2ab-2bc-2ca

(a−b−c)2=a^2+b^2+c^2−2ab+2bc−2ca

bài 2: Cho \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)

CMR: a = b = c = 1

Answer 1

Có: \( a^2+b^2+c^2+3=2(a+b+c) \)

\( =>a^2+b^2+c^2+3=2a+2b+2c \)

\( =>a^2+b^2+c^2+3-2a-2b-2c=0 \)

\(\Rightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2b+1\right)=0\)\(\Rightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)

Vì  \(\left\{{}\begin{matrix}\left(a-1\right)^2\ge0\\\left(b-1\right)^2\ge0\\\left(c-1\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\\\left(c-1\right)^2=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a-1=0\\b-1=0\\c-1=0\end{matrix}\right.\)

\(\Rightarrow a=1,b=1,c=1\)