6.178. $\frac{\sqrt{(a-x)^{2}}+\sqrt{(a-x)(b-x)}+\sqrt{(b-x)^{2}}}{\sqrt{(a-x)^{2}}-\sqrt{(a-x)(b-x)}+\sqrt{(b-x)^{2}}}=\frac{7}{3}$.
## Solution.

Domain of definition: $(a-x)(b-x)>0$.

Rewrite the original equation in the following form

$$
\begin{aligned}
& \frac{3 \sqrt{(a-x)^{2}}+3 \sqrt{(a-x)(b-x)}+3 \sqrt{(b-x)^{2}}}{3\left(\sqrt{(a-x)^{2}}-\sqrt{(a-x)(b-x)}+\sqrt{(b-x)^{2}}\right)}+ \\
& +\frac{-7 \sqrt{(a-x)^{2}}+7 \sqrt{(a-x)(b-x)}-7 \sqrt{(b-x)^{2}}}{3\left(\sqrt{(a-x)^{2}}-\sqrt{(a-x)(b-x)}+\sqrt{(b-x)^{2}}\right)}=0 \Leftrightarrow \\
& \Leftrightarrow-4 \sqrt{(a-x)^{2}}+10 \sqrt{(a-x)(b-x)}-4 \sqrt{(b-x)^{2}}=0 \\
& \text { when } \sqrt{(a-x)^{2}}-\sqrt{(a-x)(b-x)}+\sqrt{(b-x)^{2}} \neq 0
\end{aligned}
$$

Dividing both sides of the last equation by $-2 \sqrt{(b-x)^{2}} \neq 0$, we get $2 \sqrt{\left(\frac{a-x}{b-x}\right)^{2}}-5 \sqrt{\frac{a-x}{b-x}}+2=0$.

Let $\sqrt{\frac{a-x}{b-x}}=y$, where $y>0$. The equation in terms of $y$ becomes $2 y^{2}-5 y+2=0$, from which we find $y_{1}=\frac{1}{2}, y_{2}=2$. Thus, $\sqrt{\frac{a-x}{b-x}}=\frac{1}{2}$ or $\sqrt{\frac{a-x}{b-x}}=2$. This means $\frac{a-x}{b-x}=\frac{1}{4}$ or $\frac{a-x}{b-x}=4$, from which we find $x_{1}=\frac{4 a-b}{3}, x_{2}=\frac{4 b-a}{3}$.

Answer: $x_{1}=\frac{4 a-b}{3}, x_{2}=\frac{4 b-a}{3}$; if $a=b$, then there are no solutions.