6.032. $\sqrt{x+\sqrt{x+11}}+\sqrt{x-\sqrt{x+11}}=4$.

6.032. $\sqrt{x+\sqrt{x+11}}+\sqrt{x-\sqrt{x+11}}=4$.
Solution.

Let $\sqrt{x+11}=y \geq 0$ or $x+11=y^{2}$, i.e., $x=y^{2}-11$. Then

$$
\sqrt{y^{2}+y-11}+\sqrt{y^{2}-y-11}=4 \text { or } \sqrt{y^{2}+y-11}=4-\sqrt{y^{2}-y-11}
$$

Squaring both sides of the equation, we get

$$
y^{2}+y-11=16-8 \sqrt{y^{2}-y-11}+y^{2}-y-11, 8 \sqrt{y^{2}-y-11}=16-2 y
$$

or $4 \sqrt{y^{2}-y-11}=8-y$. Squaring both sides of the equation again, we find

$$
\begin{aligned}
& 16 y^{2}-16 y-176=64-16 y+y^{2}, 0<y \leq 8 \Rightarrow \\
& \Rightarrow\left\{\begin{array}{l}
15 y^{2}=240, \\
0<y \leq 8
\end{array} \text { or } y=4 .\right.
\end{aligned}
$$

From this, we get $\sqrt{x+11}=4$ or $x+11=16, x=5$. Checking, we confirm that this is a root of the original equation.

Answer: $x=5$.