https://file.aiquickdraw.com/mathbot/file/1725990436495mv6zwnl9.jpg

The expression in the image you've provided appears to be:

\[
\frac{18 \cdot N \cdot 2^N}{6 \cdot N \cdot 8 \cdot 9 \cdot \left(\frac{1}{3}\right)^{-N}}
\]

To simplify this expression, we can break it down step by step.

1. **Simplifying the numerator and denominator:**

   - The numerator is \( 18 \cdot N \cdot 2^N \).
   - The denominator is \( 6 \cdot N \cdot 8 \cdot 9 \cdot \left(\frac{1}{3}\right)^{-N} \).

2. **Cancelling common terms:**
   - The \( N \) in the numerator and denominator cancels out (as long as \( N \neq 0 \)).
   - We can also simplify the constants:
     \[
     \frac{18}{6 \cdot 8 \cdot 9} = \frac{18}{432} = \frac{1}{24}.
     \]
     So, the fraction of constants reduces to \( \frac{1}{24} \).

3. **Simplifying powers:**
   - We now have \( \frac{1}{24} \cdot \frac{2^N}{\left(\frac{1}{3}\right)^{-N}} \).
   - Using the property of exponents \( \left(\frac{1}{3}\right)^{-N} = 3^N \), we get:
     \[
     \frac{1}{24} \cdot \frac{2^N}{3^N} = \frac{1}{24} \cdot \left(\frac{2}{3}\right)^N.
     \]

Thus, the simplified expression is:

\[
\frac{1}{24} \cdot \left(\frac{2}{3}\right)^N.
\]

Let me know if you have any questions or need further clarification!

### Relative Questions:
1. What happens to the expression when \( N = 0 \)?
2. Can this expression be further simplified for specific values of \( N \)?
3. How does the term \( \left(\frac{2}{3}\right)^N \) behave as \( N \) increases?
4. What role does the constant \( \frac{1}{24} \) play in the final expression?
5. Can this form be used for calculating limits as \( N \) approaches infinity?

### Tip:
When simplifying expressions with exponents, always apply exponent rules carefully, especially when dealing with negative exponents or fractions.

Simplify the expression: \( \frac{18 \cdot N \cdot 2^N}{6 \cdot N \cdot 8 \cdot 9 \cdot \left(\frac{1}{3}\right)^{-N}} \).

Simplify the algebraic expression involving exponents and rational fractions: \( \frac{18 \cdot N \cdot 2^N}{6 \cdot N \cdot 8 \cdot 9 \cdot \left(\frac{1}{3}\right)^{-N}} \). Learn the step-by-step solution including exponent rules and fraction simplification. Suitable for Grades 9-12.

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