\hfill\break 根据偶次方根下非负，且分母不为零，列式即可得解.\\ 【详解】\\ 由$f(x)=\frac{\sqrt{2x-1}}{{{x}^{2}}-1}$，\\ 可得：$\left { \begin{align} \& 2x-1\ge 0    \& {{x}^{2}}-1
e 0    \end{align} \right.$，解得：$x\ge \frac{1}{2}$且$x
e 1$，\\ 故选：D. 
函数$f(x)=\frac{\sqrt{2x-1}}{{{x}^{2}}-1}$的定义城为（~~~~~~~）

$[\frac{1}{2},+\infty )$ 
(1+∞) 
$\text{ } ! !\textasciitilde ! !\text{ (-1,}\frac{\text{1}}{\text{2}}\text{)}\cup \text{(1,+}\infty \text{) } ! !\textasciitilde ! !\text{ }$ 
$[\frac{\text{1}}{\text{2}}\text{,1)U(1,+}\infty \text{)}$ 