$1 \cdot 73$ The value of the algebraic expression $\sqrt{11+6 \sqrt{2}}+\sqrt{11-6 \sqrt{2}}$ is
(A) a natural number.
(B) an irrational number.
(C) a fraction.
(D) none of the above conclusions is correct.
(2nd China Tianjin "Chinese Youth Cup" Mathematics Invitational, 1988)
[Solution 1] $(\sqrt{11+6 \sqrt{2}}+\sqrt{11-6 \sqrt{2}})^{2}$
$$
\begin{aligned}
&=11+6 \sqrt{2}+2 \sqrt{11^{2}-(6 \sqrt{2})^{2}}+11-6 \sqrt{2} \\
&=22+2 \sqrt{49}=36,
\end{aligned}
$$
$\therefore \quad$ the original expression $=6$.
Therefore, the answer is (A).
[Solution 2] Notice the following transformation:
$$
\begin{aligned}
\text { the original expression } & =\sqrt{9+2 \cdot 3 \cdot \sqrt{2}+2}+\sqrt{9-2 \cdot 3 \cdot \sqrt{2}+2} \\
& =\sqrt{(3+\sqrt{2})^{2}}+\sqrt{(3-\sqrt{2})^{2}} \\
& =(3+\sqrt{2})+(3-\sqrt{2})=6 .
\end{aligned}
$$

Therefore, the answer is $(A)$.