$\text{ Number of integers satisfying }\sqrt{x-6}-\sqrt{10-x}\ge 1\text{ is }$

$\begin{array}{r}\text{ We have }\sqrt{x-6}-\sqrt{10-x}\ge 1\\ \text{ We must have }x\ge 6\text{ and }x\le 10\end{array}$

$\begin{array}{r}\text{ Then }\sqrt{x-6}\ge 1+\sqrt{10-x}\\ \text{ Squaring given inequality, }\end{array}$

$\begin{array}{l}x-6\ge 1+10-x+2\sqrt{10-x}\\ \Rightarrow 2x-17\ge 2\sqrt{10-x}\end{array}$

Clearly LHS must be positive.

$\therefore x\ge \frac{17}{2}$

so, we have

$\begin{array}{l}\begin{array}{rl} & 4x^2+289-68x\ge 4(10-x)\\ \Rightarrow & 4x^2-64x+249\ge 0\\ \Rightarrow & x^2-16x+\frac{249}{4}\ge 0\\ \Rightarrow & (x-8{)}^2-{\left(\frac{\sqrt{7}}{2}\right)}^2\ge 0\\ \Rightarrow & \left(x-8-\frac{\sqrt{7}}{2}\right)\left(x-8+\frac{\sqrt{7}}{2}\right)\ge 0\end{array}\\ \mathrm{From} \mathrm{the} \mathrm{sign} \mathrm{scheme}, x\in \left[\right(16+\sqrt{7})/2,10]\end{array}$