The problem asks us to simplify the following expression:

$\frac{\sqrt{x(x^{2n+1})}}{\sqrt[3]{x^{3n}}}, \quad x > 0.$

Step-by-Step Simplification:

1. Simplify the numerator: The numerator is: $\sqrt{x(x^{2n+1})} = \sqrt{x^{1} \cdot x^{2n+1}} = \sqrt{x^{2n+2}}.$ Since $x > 0$, we can simplify: $\sqrt{x^{2n+2}} = x^{\frac{2n+2}{2}} = x^{n+1}.$

2. Simplify the denominator: The denominator is: $\sqrt[3]{x^{3n}} = (x^{3n})^{1/3} = x^{3n \cdot \frac{1}{3}} = x^{n}.$

3. Combine the numerator and denominator: The full expression becomes: $\frac{x^{n+1}}{x^{n}}.$ Using the property of exponents $\frac{x^a}{x^b} = x^{a-b}$, we simplify: $x^{n+1 - n} = x^{1}.$

Final Answer:

$\boxed{x}$

Would you like further clarification or more details on any step?

Here are 5 related questions to expand your understanding:

1. How do you simplify expressions involving roots and exponents?
2. Why can $\sqrt{x^{2n+2}}$ be rewritten as $x^{n+1}$ when $x > 0$?
3. What is the difference between a square root and a cube root in simplifications?
4. Can you apply these principles to simplify $\frac{\sqrt{x^{5}}}{\sqrt[3]{x^{6}}}$?
5. How does the condition $x > 0$ simplify working with exponents and radicals?

Tip: Always check for restrictions on the variable (like $x > 0$) to ensure your simplifications are valid!