6.050. $\frac{(5-x) \sqrt{5-x}+(x-3) \sqrt{x-3}}{\sqrt{5-x}+\sqrt{x-3}}=2$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}5-x \geq 0, \\ x-3 \geq 0,\end{array} \Leftrightarrow 3 \leq x \leq 5\right.$.

Rewrite the equation as $\frac{(\sqrt{5-x})^{3}+(\sqrt{x-3})^{3}}{\sqrt{5-x}+\sqrt{x-3}}=2$ and factor the numerator of the left side as the sum of cubes:

$$
\frac{(\sqrt{5-x}+\sqrt{x-3})\left(\sqrt{(5-x)^{2}}-\sqrt{(5-x)(x-3)}+\sqrt{(x-3)^{2}}\right)}{\sqrt{5-x}+\sqrt{x-3}}=2
$$

Considering that the denominator is positive, we get

$$
\begin{aligned}
& \sqrt{(5-x)^{2}}-\sqrt{(5-x)(x-3)}+\sqrt{(x-3)^{2}}=2 \Leftrightarrow \\
& \Leftrightarrow 5-x-\sqrt{(5-x)(x-3)}+x-3=2 \Leftrightarrow \\
& \Leftrightarrow \sqrt{(5-x)(x-3)}=0,(5-x)(x-3)=0,
\end{aligned}
$$

from which $5-x=0, x_{1}=5$, or $x-3=0, x_{2}=3$. By verification, we confirm that these are the roots of the original equation.

Answer: $x_{1}=5, x_{2}=3$.