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

Therefore, the original expression
$$
\begin{array}{l}
=\sqrt{4 a(9 a+2)+10 a+1-6 a} \\
=\sqrt{(6 a+1)^{2}}=6 a+1=\underbrace{66 \cdots 6}_{n-1 \uparrow} .
\end{array}
$$