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The problem requires us to simplify the given expression so that it is in the form \( K x^a y^b z^c \).

The expression is:

\[
z^{-4} \left( \frac{7x^2 y^3}{z y^2 \sqrt{5}} \right)^2 \cdot 12^0
\]

### Step-by-Step Simplification:

1. **Handle \( 12^0 \):**
   \[
   12^0 = 1
   \]
   So the expression simplifies to:
   \[
   z^{-4} \left( \frac{7x^2 y^3}{z y^2 \sqrt{5}} \right)^2
   \]

2. **Square the fraction inside the parentheses:**
   \[
   \left( \frac{7x^2 y^3}{z y^2 \sqrt{5}} \right)^2 = \frac{(7^2)(x^2)^2(y^3)^2}{(z^2)(y^2)^2(\sqrt{5})^2}
   \]
   Simplifying:
   \[
   = \frac{49 x^4 y^6}{z^2 y^4 \cdot 5}
   \]
   So the expression becomes:
   \[
   z^{-4} \cdot \frac{49 x^4 y^6}{5 z^2 y^4}
   \]

3. **Combine terms:**
   The expression now is:
   \[
   \frac{49 x^4 y^6}{5 z^6 y^4}
   \]
   Simplify the powers of \( y \) and \( z \):
   \[
   = \frac{49 x^4 y^2}{5 z^6}
   \]

Now, we have the expression in the form \( K x^a y^b z^c \). Comparing the simplified expression with \( K x^a y^b z^c \), we get:

- \( K = \frac{49}{5} \)
- \( a = 4 \)
- \( b = 2 \)
- \( c = -6 \)

### Final Answers:
- \( a = 4 \)
- \( b = 2 \)
- \( c = -6 \)
- \( K = \frac{49}{5} \)

Let me know if you need further details or have any questions.

### Related Questions:
1. How do we simplify powers of variables in fractions?
2. What is the general rule for exponents when squaring a fraction?
3. Why is \( 12^0 \) equal to 1?
4. How do square roots behave when raised to even powers?
5. How do negative exponents affect variable expressions?

**Tip:** Always simplify powers and constants separately before combining terms!

Simplify the following expression as much as possible so that the final expression is in the form Kx^a y^b z^c: z^{-4} * (7x^2 y^3 / (z y^2 √5))^2 * 12^0

Learn how to simplify the expression z^{-4} * (7x^2 y^3 / (z y^2 √5))^2 * 12^0 with a detailed step-by-step solution. This problem explores key algebra concepts like negative exponents, the power rule, and fractional powers, making it ideal for high school students.

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