$6.299 \frac{2}{19}\left(\sqrt{x^{2}+37 x+336}-\sqrt{x^{2}+18 x+32}\right)=\sqrt{\frac{21+x}{16+x}}$.

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$6.299 \frac{2}{19}\left(\sqrt{x^{2}+37 x+336}-\sqrt{x^{2}+18 x+32}\right)=\sqrt{\frac{21+x}{16+x}}$.
Solution. Note that $x^{2}+37 x+336 \geq x^{2}+18 x+32$, i.e., $x \geq-16$.

Therefore, the original equation is equivalent to

$$
\begin{aligned}
& \left\{\begin{array}{l}
\frac{2}{19} \sqrt{x+16}(\sqrt{x+21}-\sqrt{x+2})=\frac{\sqrt{x+21}}{\sqrt{x+16}}, \Leftrightarrow \\
x>-16
\end{array}\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
\frac{2(x+16)(\sqrt{x+21}-\sqrt{x+2})}{(x+21)-(x+2)}=\sqrt{x+21}, \Leftrightarrow \frac{2(x+16)}{\sqrt{x+21}+\sqrt{x+2}}=\sqrt{x+21} \Leftrightarrow \\
x>-16
\end{array}\right. \\
& \Leftrightarrow 2(x+16)=x+21+\sqrt{x+21} \cdot \sqrt{x+2} \Leftrightarrow\left\{\begin{array}{l}
(x+11)^{2}=(x+21)(x+2), \\
x \geq-2
\end{array} \Leftrightarrow x=79\right.
\end{aligned}
$$

Answer: $x=79$.

$$