Question

[Nguyễn Quốc Khánh]

Nếu : \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)  thì a=b=c=1

Answer 1

[zZz Cool Kid_new zZz]

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

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

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

Dấu "=" xảy ra khi \(a=b=c=1\)

Answer 2

[Huyền Nhi]

Ta có: \(a^2+b^2+c^2+3=2.\left(a+b+c\right)\)

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

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

\(\Rightarrow\hept{\begin{cases}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\\\left(c-1\right)^2=0\end{cases}}\text{vì }\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\)

\(\Leftrightarrow a=b=c=1\)

Vậy a=b=c=1