6.173. $2 \sqrt{5 \sqrt[4]{x+1}+4}-\sqrt{2 \sqrt[4]{x+1}-1}=\sqrt{20 \sqrt[4]{x+1}+5}$.
Solution.

Domain of definition: $\left\{\begin{array}{l}x+1 \geq 0, \\ 2 \sqrt[4]{x+1}-1 \geq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}x \geq-1, \\ 2 \sqrt[4]{x+1} \geq 1,\end{array} \Leftrightarrow x \geq-\frac{15}{16}\right.\right.$.

Let $\sqrt[4]{x+1}=y$, where $y \geq 0$. The equation in terms of $y$ becomes $2 \sqrt{5 y+4}-\sqrt{2 y-1}=\sqrt{20 y+5}$. Squaring both sides of the equation, we get

$$
\begin{aligned}
& 20 y+16-4 \sqrt{(5 y+4)(2 y-1)}+2 y-1=20 y+5 \Leftrightarrow \\
& \Leftrightarrow y+5=2 \sqrt{(5 y+4)(2 y-1)} \Leftrightarrow y^{2}+10 y+25=4\left(10 y^{2}+3 y-4\right) \Leftrightarrow \\
& \Leftrightarrow 39 y^{2}+2 y-41=0, \text { from which } y_{1}=-\frac{41}{39}, y_{2}=1 ; y_{1}=-\frac{41}{39}<0 \text { is not suitable. }
\end{aligned}
$$

Hence, $\sqrt[4]{x+1}=1 \Leftrightarrow x+1=1, x=0$.

Answer: $x=0$.