$\sqrt{5+\sqrt{x-1}}=6-x$ (Đk : $1 \leq x \leq 6$)$\Leftrightarrow 5+\sqrt{x-1}=x^2-12x+36$
$\Leftrightarrow x^2-11x+26=\sqrt{x-1}+x-5$
$\Leftrightarrow x^2-11x+26=\frac{-x^2+11x-26}{\sqrt{x-1}+5-x}$
$\Leftrightarrow(x^2-11x+26)(1+\frac{1}{\sqrt{x-1}+5-x})=0$
giải ra chỉ có $\color{blue}{x= \frac{11-\sqrt{17}}{2}}$ thỏa mãn đk