(9) Calculate $\sqrt{\underbrace{44 \cdots 4}_{2 n \text { digits }}+\underbrace{11 \cdots 1}_{n+1 \text { digits }}-\underbrace{66 \cdots 6}_{n \text { digits }}}=$
Let $\underbrace{11 \cdots 1}_{n \text { digits }}=a$, then
$$
\begin{array}{c}
10^{n}=\underbrace{99 \cdots 9}_{n \text { digits }}+1=9 a+1, \\
\underbrace{44 \cdots 4}_{2 n \text { digits }}=4 a \cdot 10^{n}+4 a=4 a(9 a+2), \\
\underbrace{11 \cdots 1}_{n+1 \text { digits }}=10 a+1, \\
\underbrace{66 \cdots 6}_{n \text { digits }}=6 a,
\end{array}
$$

Therefore, the original expression equals
$$
\sqrt{4 a(9 a+2)+10 a+1-6 a}=\sqrt{(6 a+1)^{2}}=6 a+1=\underbrace{66 \cdots 67}_{n-1 \text { digits }}.
$$

Thus, the answer is $\underbrace{66 \cdots 67}_{n-1 \text { digits }}$.