Solution:
Any three numbers in $[5,10]$.

$\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1$

can be written

$\sqrt{(x-1)-4 \sqrt{x-1}+4}+\sqrt{(x-1)-6 \sqrt{x-1}+9}=\sqrt{(\sqrt{x-1}-2)^{2}}+\sqrt{(\sqrt{x-1}-3)^{2}}=1$

hence $|\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1$.

There are four cases.

(I) $\sqrt{x-1}-2 \geq 0$ and $\sqrt{x-1}-3 \geq 0$ which implies $x \geq 10 \Rightarrow x=10$.

(II) $\sqrt{x-1}-2 \leq 0$ and $\sqrt{x-1}-3 \leq 0$ which implies $x \leq 5 \Rightarrow x=5$.

(III) $\sqrt{x-1}-2 \geq 0$ and $\sqrt{x-1}-3 \leq 0$ implies $5 \leq x \leq 10$.

(IV) $\sqrt{x-1}-2 \leq 0$ and $\sqrt{x-1}-3 \geq 0$, which is impossible.

Thus, any $x \in[5,10]$ solves the equation.