c/m: a^2+b^2+c^2=ab+ac+bc thì a=b=c a^2+b^2+c^2+3=2(a+b+c) thì a=b=c=1

Giải thích các bước giải:

 a) `a^2+b^2+c^2=ab+ac+bc`

`=>2a^2+2b^2+2c^2=2ab+2ac+2bc`

`=>2a^2+2b^2+2c^2-2ab-2ac-2bc=0`

`=>(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ac+a^2)=0`

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

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

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

Dấu `=` xảy ra `<=>`$\left\{\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.$`=>`$\left\{\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.$`=>a=b=c`

 b) `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`

`=>(a^2-2a+1)+(b^2-2b+1)+(c^2-2c+1)=0`

`=>(a-1)^2+(b-1)^2+(c-1)^2=0`

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

`=>(a-1)^2+(b-1)^2+(c-1)^2>=0`

Dấu `=` xảy ra `<=>`$\left\{\begin{matrix}a-1=0\\b-1=0\\c-1=0\end{matrix}\right.$`=>`$\left\{\begin{matrix}a=1\\b=1\\c=1\end{matrix}\right.$`=>a=b=c=1`