`~rai~`
\(a)x^2+y^2+z^2=xy+yz+zx\\\Leftrightarrow 2x^2+2y^2+2z^2=2xy+2yz+2xz\\\Leftrightarrow 2x^2+2y^2+2z^2-2zy-2yz-2xz=0\\\Leftrightarrow (x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)=0\\\Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2=0\\\text{Ta có:}(x-y)^2\ge 0;(y-z)^2\ge 0;(z-x)^2\ge 0\quad\forall x;y;z\\\Rightarrow (x-y)^2+(y-z)^2+(z-x)^2\ge 0\quad\forall x;y;z\\\text{mà }(x-y)^2+(y-z)^2+(z-x)^2=0\\\Rightarrow \begin{cases}(x-y)^2=0\\(y-z)^2=0\\(z-x)^2=0\end{cases}\\\Leftrightarrow \begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}\\\Leftrightarrow \begin{cases}x=y\\y=z\\z=x\end{cases}\\\Leftrightarrow x=y=z\\\text{mà }x+y+z=3\\\Rightarrow x=y=z=1.\\\text{Vậy } x=y=z=1.\\b)x^2+y^2=x+y-\dfrac{1}{2}\\\Leftrightarrow x^2+y^2-x-y+\dfrac{1}{2}=0\\\Leftrightarrow \left(x^2-x+\dfrac{1}{4}\right)+\left(y^2-y+\dfrac{1}{4}\right)=0\\\Leftrightarrow \left(x-\dfrac{1}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=0\\\text{Ta có:}\left(x-\dfrac{1}{2}\right)^2\ge 0;\left(y-\dfrac{1}{2}\right)^2\ge 0\quad\forall x;y\\\Rightarrow \left(x-\dfrac{1}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2\ge 0\quad\forall x;y\\\text{mà }\left(x-\dfrac{1}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=0\\\Rightarrow \begin{cases}\left(x-\dfrac{1}{2}\right)^2=0\\\left(y-\dfrac{1}{2}\right)^2=0\end{cases}\\\Leftrightarrow \begin{cases}x-\dfrac{1}{2}=0\\y-\dfrac{1}{2}=0\end{cases}\\\Leftrightarrow \begin{cases}x=\dfrac{1}{2}\\y=\dfrac{1}{2}.\end{cases}\\\text{Vậy }x=y=\dfrac{1}{2}.\)
