###### tags: `數與式`
# 挑戰題 8.

試求有序數對 $(x,y,z)=\underline{\qquad\qquad}$ ，滿足 $\begin{align}\sqrt{x+\frac{1}{2}}+\sqrt{y-4}+\sqrt{z^2+1}=\frac{1}{2}(x+y+z^2+\frac{1}{2})\end{align}$

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$\boxed{答}：\begin{align}(\frac{1}{2},5,0)\end{align}$
$\boxed{解}：$ 令 $\begin{align}a=\sqrt{x+\frac{1}{2}}\end{align}$，$b=\sqrt{y-4}$，$c=\sqrt{z^2+1}$
$\begin{align}\Rightarrow x=a^2-\frac{1}{2}，y=b^2+4，z^2=c^2-1\end{align}$
$\begin{align}a+b+c=\frac{1}{2}(a^2+b^2+c^2+3)\end{align}$
$a^2+b^2+c^2-2a-2b-2c+3=0$
配方 $(a-1)^2+(b-1)^2+(c-1)^2=0$
$\Rightarrow a=b=c=1$
$\begin{align}\Rightarrow x=\frac{1}{2}，y=5，z=0\end{align}$