1. Let the real number
$$
x=\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}} \text {. }
$$

Then $\sqrt{2009 x}=$ $\qquad$
$$
\begin{array}{l} 
\text { Two } \\
x^{3}=(\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}})^{3} \\
=40+6(\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}) \\
=40+6 x,
\end{array}
$$

From $x^{3}=(\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}})^{3}$

That is,
$$
\begin{array}{l}
x^{3}-6 x-40=0 \\
\Rightarrow(x-4)\left(x^{2}+4 x+10\right)=0 .
\end{array}
$$

Clearly, $x^{2}+4 x+10>0$.
So, $x=4$.
$$
\begin{array}{l}
\text { Therefore } \sqrt{2009 x}=\sqrt{2009 \times 4} \\
=\sqrt{41 \times 49 \times 4}=14 \sqrt{41} .
\end{array}
$$