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We are given the equation:

\[
\left\lfloor \sqrt{n} + \sqrt{n+1} + \sqrt{n+2} \right\rfloor = \left\lfloor \sqrt{9n+8} \right\rfloor
\]

where \(\lfloor x \rfloor\) represents the floor function, which gives the greatest integer less than or equal to \(x\).

### Steps to Solve:

1. **Approximation for Large \(n\)**:
   We need to compare the sum of the square roots on the left-hand side with the square root on the right-hand side for different values of \(n\). Since both expressions are within a floor function, we can start by calculating them for small values of \(n\).

2. **Testing for Small \(n\):**

   Let's try substituting small values of \(n\) and check both sides.

   - For \(n = 0\):
     \[
     \text{LHS} = \lfloor \sqrt{0} + \sqrt{1} + \sqrt{2} \rfloor = \lfloor 0 + 1 + 1.414 \rfloor = \lfloor 2.414 \rfloor = 2
     \]
     \[
     \text{RHS} = \lfloor \sqrt{9(0) + 8} \rfloor = \lfloor \sqrt{8} \rfloor = \lfloor 2.828 \rfloor = 2
     \]
     So, the equation holds for \(n = 0\).

   - For \(n = 1\):
     \[
     \text{LHS} = \lfloor \sqrt{1} + \sqrt{2} + \sqrt{3} \rfloor = \lfloor 1 + 1.414 + 1.732 \rfloor = \lfloor 4.146 \rfloor = 4
     \]
     \[
     \text{RHS} = \lfloor \sqrt{9(1) + 8} \rfloor = \lfloor \sqrt{17} \rfloor = \lfloor 4.123 \rfloor = 4
     \]
     The equation holds for \(n = 1\) as well.

   - For \(n = 2\):
     \[
     \text{LHS} = \lfloor \sqrt{2} + \sqrt{3} + \sqrt{4} \rfloor = \lfloor 1.414 + 1.732 + 2 \rfloor = \lfloor 5.146 \rfloor = 5
     \]
     \[
     \text{RHS} = \lfloor \sqrt{9(2) + 8} \rfloor = \lfloor \sqrt{26} \rfloor = \lfloor 5.099 \rfloor = 5
     \]
     The equation holds for \(n = 2\).

   - For \(n = 3\):
     \[
     \text{LHS} = \lfloor \sqrt{3} + \sqrt{4} + \sqrt{5} \rfloor = \lfloor 1.732 + 2 + 2.236 \rfloor = \lfloor 5.968 \rfloor = 5
     \]
     \[
     \text{RHS} = \lfloor \sqrt{9(3) + 8} \rfloor = \lfloor \sqrt{35} \rfloor = \lfloor 5.916 \rfloor = 5
     \]
     The equation holds for \(n = 3\).

   So far, the equation holds for \(n = 0, 1, 2, 3\).

3. **Conclusion**:
   Based on the calculations, the equation appears to be true for the tested values of \(n\), and it likely holds for larger values of \(n\) as well. 

Would you like a more in-depth analysis of how the function behaves for larger \(n\) or any specific questions related to this solution? Let me know if you need further details!

Here are 5 related questions to think about:
1. What is the general behavior of square root functions for large \(n\)?
2. How does the floor function affect approximations for continuous functions like square roots?
3. What other mathematical functions behave similarly to square roots when approximated for small or large values?
4. Can we generalize this kind of equation for other sequences of square roots?
5. What would happen if we replaced the floor function with a ceiling function (\(\lceil x \rceil\))?

**Tip**: When solving equations involving square roots and the floor function, always check small values first to see if a pattern emerges!

Solve the equation: ⌊√n + √(n+1) + √(n+2)⌋ = ⌊√(9n+8)⌋

This problem involves solving an equation that includes floor functions and square roots. It explores the properties of square roots and how they behave when combined with floor functions. The solution is suitable for high school students in Grades 9-12.

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