The set of values of x satisfying the in equation $\sqrt{{\mathrm{x}}^2-18\mathrm{x}+72}<\mathrm{x}-1$is 

Given $\sqrt{{\mathrm{x}}^2-18\mathrm{x}+72}<\mathrm{x}-1$

To define the above inequality $x^2-18x+72\ge 0\Rightarrow \left(x-12\right)\left(x-6\right)\ge 0\Rightarrow x\le 6 or x\ge 12$

Squaring on both sides

$\begin{array}{rcl}{x}^2-18x+72& <& x^2-2x+1\\ -16x& <& -71\\ x& >& \frac{71}{16}\end{array}$

Therefore, the solution set for the inequality $\sqrt{{\mathrm{x}}^2-18\mathrm{x}+72}<\mathrm{x}-1$ is $x\in \boxed{[12,\infty )}$