9,10 |
| :---: | :---: | :---: |
|  | [ Formulas for abbreviated multiplication (other).] |  |
|  | [ Properties of the modulus. Triangle inequality ] |  |
|  | equations with moduli |  |

Solve the equation: $\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1$.
Note that $x+3-4 \sqrt{x-1}=(\sqrt{x-1}-2)^{2}, x+8-6 \sqrt{x-1}=(\sqrt{x-1}-3)^{2}$.

Therefore, the original equation can be written as $|\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1$.

By the triangle inequality, $|t-2|+|t-3| \geq 3-2=1$, and equality holds only for points on the segment [2, 3]. Thus, $2 \leq \sqrt{x-1} \leq 3$, which means $5 \leq x \leq 10$.

## Answer

$5 \leq x \leq 10$.