6.170. $\frac{\sqrt{x+4}+\sqrt{x-4}}{2}=x+\sqrt{x^{2}-16}-6$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}x+4 \geq 0, \\ x-4 \geq 0,\end{array} \Leftrightarrow x \geq 4\right.$.

By squaring both sides of the equation, we get

$$
\begin{aligned}
& \frac{x+4+2 \sqrt{(x+4)(x-4)}+x-4}{4}=\left(x+\sqrt{x^{2}-16}\right)^{2}-12\left(x+\sqrt{x^{2}-16}\right)+36 \\
& 2\left(x+\sqrt{x^{2}-16}\right)-25\left(x+\sqrt{x^{2}-16}\right)+72=0
\end{aligned}
$$

Let $x+\sqrt{x^{2}-16}=y$. The equation in terms of $y$ becomes

$$
2 y^{2}-25 y+72=0, \text { from which } y_{1}=\frac{9}{2}, y_{2}=8
$$

Then

$$
\text { 1) } x+\sqrt{x^{2}-16}=\frac{9}{2} \Leftrightarrow 2 \sqrt{x^{2}-16}=9-2 x, \text { where } 9-2 x \geq 0, x \leq 4.5
$$

After squaring the equation, we get

$$
4 x^{2}-64=81-36 x+4 x^{2}, 36 x=145, x_{1}=\frac{145}{36} ; \text { does not fit; }
$$

2) $x+\sqrt{x^{2}-16}=8, \sqrt{x^{2}-16}=8-x$, where $8-x \geq 0, x \leq 8$. After squaring the equation, we get $x^{2}-16=64-16 x+x^{2}, x=5$.

Answer: $x=5$.