1. **State the problem:** Simplify the function $$f(x) = \frac{\sqrt{x-6} + \sqrt{2x-2}}{\sqrt{x-1} - \sqrt{x-5}}.$$\n\n2. **Identify the domain:** For the square roots to be defined, the radicands must be non-negative:\n- $x-6 \geq 0 \Rightarrow x \geq 6$\n- $2x-2 \geq 0 \Rightarrow x \geq 1$\n- $x-1 \geq 0 \Rightarrow x \geq 1$\n- $x-5 \geq 0 \Rightarrow x \geq 5$\n\nThe most restrictive domain is $x \geq 6$.\n\n3. **Simplify the denominator:** Use the conjugate to rationalize the denominator:\n$$\sqrt{x-1} - \sqrt{x-5} = \frac{(\sqrt{x-1} - \sqrt{x-5})(\sqrt{x-1} + \sqrt{x-5})}{\sqrt{x-1} + \sqrt{x-5}} = \frac{(x-1) - (x-5)}{\sqrt{x-1} + \sqrt{x-5}} = \frac{4}{\sqrt{x-1} + \sqrt{x-5}}.$$\n\n4. **Rewrite the function:**\n$$f(x) = \frac{\sqrt{x-6} + \sqrt{2x-2}}{\sqrt{x-1} - \sqrt{x-5}} = \left(\sqrt{x-6} + \sqrt{2x-2}\right) \cdot \frac{\sqrt{x-1} + \sqrt{x-5}}{4}.$$\n\n5. **Simplify the numerator:** Note that $\sqrt{2x-2} = \sqrt{2(x-1)} = \sqrt{2} \sqrt{x-1}$. So numerator becomes:\n$$\left(\sqrt{x-6} + \sqrt{2} \sqrt{x-1}\right) \left(\sqrt{x-1} + \sqrt{x-5}\right).$$\n\n6. **Expand the numerator:**\n$$\sqrt{x-6} \cdot \sqrt{x-1} + \sqrt{x-6} \cdot \sqrt{x-5} + \sqrt{2} \sqrt{x-1} \cdot \sqrt{x-1} + \sqrt{2} \sqrt{x-1} \cdot \sqrt{x-5}.$$\n\nSimplify terms:\n- $\sqrt{x-6} \cdot \sqrt{x-1} = \sqrt{(x-6)(x-1)}$\n- $\sqrt{x-6} \cdot \sqrt{x-5} = \sqrt{(x-6)(x-5)}$\n- $\sqrt{2} \sqrt{x-1} \cdot \sqrt{x-1} = \sqrt{2} (x-1)$\n- $\sqrt{2} \sqrt{x-1} \cdot \sqrt{x-5} = \sqrt{2} \sqrt{(x-1)(x-5)}$\n\n7. **Final expression:**\n$$f(x) = \frac{1}{4} \left[ \sqrt{(x-6)(x-1)} + \sqrt{(x-6)(x-5)} + \sqrt{2} (x-1) + \sqrt{2} \sqrt{(x-1)(x-5)} \right].$$\n\nThis is the simplified form of $f(x)$ valid for $x \geq 6$.