Let's break down the expression step by step:

The expression is:

\[
4^0 - \left(\frac{3}{4} + \frac{4}{7}\right) - 81^2
\]

1. **Calculate \(4^0\):**

   \[
   4^0 = 1
   \]

2. **Add the fractions:**

   \[
   \frac{3}{4} + \frac{4}{7}
   \]

   To add these fractions, find a common denominator, which is 28:

   \[
   \frac{3}{4} = \frac{3 \times 7}{4 \times 7} = \frac{21}{28}
   \]
   \[
   \frac{4}{7} = \frac{4 \times 4}{7 \times 4} = \frac{16}{28}
   \]

   Now, add them together:

   \[
   \frac{21}{28} + \frac{16}{28} = \frac{37}{28}
   \]

3. **Calculate \(81^2\):**

   \[
   81^2 = 6561
   \]

Now, plug these values back into the expression:

\[
1 - \frac{37}{28} - 6561
\]

4. **Simplify the expression:**

   Convert \(1\) to a fraction with the same denominator:

   \[
   1 = \frac{28}{28}
   \]

   Subtract the fractions:

   \[
   \frac{28}{28} - \frac{37}{28} = \frac{28 - 37}{28} = \frac{-9}{28}
   \]

   So the expression becomes:

   \[
   \frac{-9}{28} - 6561
   \]

5. **Final Result:**

   Since \(\frac{-9}{28}\) is a small fraction, subtracting \(6561\) from it would give a large negative number:

   \[
   \frac{-9}{28} - 6561 = \text{approximately } -6561.32
   \]

So, the final answer is approximately \(-6561.32\).

Would you like more details or have any questions? Here are some related questions you might consider:

1. How do you find the least common denominator when adding fractions?
2. What is the general rule for any non-zero number raised to the power of zero?
3. What are some common mistakes to avoid when dealing with exponents and large numbers?
4. How can you simplify the process of subtracting fractions from whole numbers?
5. What other methods can you use to handle large exponentiation problems?

**Tip:** When dealing with complex expressions, always simplify individual components before combining them.

Solve the mathematical expression 4^0 - (3/4 + 4/7) - 81^2 step-by-step. This solution covers exponentiation rules, addition of fractions, and subtracting large numbers. Suitable for students in Grades 8-10.

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