jmc

Note that for the expression to be defined, we must have $x\ge 1.$ Let $y=\sqrt{x-1}.$ Then $y^2=x-1,$ so $x=y^2+1.$ We can then write the given equation as $$\sqrt{y^2-4y+4}+\sqrt{y^2-6y+9}=1.$$Thus, $\sqrt{(y-2{)}^2}+\sqrt{(y-3{)}^2}=1,$ or $$|y-2|+|y-3|=1.$$If $y<2,$ then $$|y-2|+|y-3|=2-y+3-y=5-2y>1.$$If $y>3,$ then $$|y-2|+|y-3|=y-2+y-3=2y-5>1.$$If $2\le y\le 3,$ then $$|y-2|+|y-3|=y-2+3-y=1,$$so we must have $2\le y\le 3.$ Then $$2\le \sqrt{x-1}\le 3,$$so $$4\le x-1\le 9,$$or $5\le x\le 10.$ Thus, the solution is $x\in \boxed{[5,10]}.$