https://file.aiquickdraw.com/mathbot/file/1726718088761xiyqu5ak.jpg

Let's tackle the problems step by step:

### 1. Simplify the expression:
\[
\left( \frac{a^{-3} b^{-2} c}{a^{-5} b^{-7} c^2} \right)^{-2}
\]

We first simplify the fraction inside the parentheses:

#### Step 1: Combine the powers
Using the rule of exponents \( \frac{x^m}{x^n} = x^{m-n} \), we simplify each variable:

- For \( a \):
  \[
  a^{-3} \div a^{-5} = a^{-3 - (-5)} = a^{2}
  \]
  
- For \( b \):
  \[
  b^{-2} \div b^{-7} = b^{-2 - (-7)} = b^{5}
  \]
  
- For \( c \):
  \[
  c \div c^2 = c^{1 - 2} = c^{-1}
  \]

Now the expression inside the parentheses becomes:
\[
(a^2 b^5 c^{-1})^{-2}
\]

#### Step 2: Apply the outer exponent \((-2)\)
Now, we distribute the exponent \((-2)\) to each term inside the parentheses:
\[
(a^2)^{-2} = a^{-4}, \quad (b^5)^{-2} = b^{-10}, \quad (c^{-1})^{-2} = c^2
\]

Thus, the final simplified expression is:
\[
a^{-4} b^{-10} c^2
\]

### 2. Find the value of:
\[
a^{\frac{1}{2}} b^{-\frac{3}{4}} c^{-\frac{2}{3}}
\]
Given: \( a = 16 \), \( b = 625 \), \( c = 27 \)

#### Step 1: Calculate \( a^{\frac{1}{2}} \):
\[
a^{\frac{1}{2}} = 16^{\frac{1}{2}} = \sqrt{16} = 4
\]

#### Step 2: Calculate \( b^{-\frac{3}{4}} \):
\[
b^{-\frac{3}{4}} = 625^{-\frac{3}{4}} = \left( 625^{\frac{1}{4}} \right)^{-3} = 5^{-3} = \frac{1}{125}
\]

#### Step 3: Calculate \( c^{-\frac{2}{3}} \):
\[
c^{-\frac{2}{3}} = 27^{-\frac{2}{3}} = \left( 27^{\frac{1}{3}} \right)^{-2} = 3^{-2} = \frac{1}{9}
\]

#### Step 4: Multiply the results:
\[
a^{\frac{1}{2}} b^{-\frac{3}{4}} c^{-\frac{2}{3}} = 4 \times \frac{1}{125} \times \frac{1}{9} = \frac{4}{1125}
\]

### 3. Rationalize the expression:
\[
\frac{2 - \sqrt{5}}{2 + \sqrt{5}}
\]

#### Step 1: Multiply the numerator and denominator by the conjugate of the denominator:
\[
\frac{2 - \sqrt{5}}{2 + \sqrt{5}} \times \frac{2 - \sqrt{5}}{2 - \sqrt{5}} = \frac{(2 - \sqrt{5})^2}{(2 + \sqrt{5})(2 - \sqrt{5})}
\]

#### Step 2: Simplify the denominator using the difference of squares formula:
\[
(2 + \sqrt{5})(2 - \sqrt{5}) = 2^2 - (\sqrt{5})^2 = 4 - 5 = -1
\]

#### Step 3: Simplify the numerator:
\[
(2 - \sqrt{5})^2 = 2^2 - 2 \times 2 \times \sqrt{5} + (\sqrt{5})^2 = 4 - 4\sqrt{5} + 5 = 9 - 4\sqrt{5}
\]

#### Step 4: Put it all together:
\[
\frac{(2 - \sqrt{5})^2}{(2 + \sqrt{5})(2 - \sqrt{5})} = \frac{9 - 4\sqrt{5}}{-1} = -9 + 4\sqrt{5}
\]

Thus, the rationalized expression is:
\[
-9 + 4\sqrt{5}
\]

---

Would you like further details on any step? Here are some related questions you might find useful:

1. How do you handle more complex expressions with negative exponents?
2. What other methods can be used to simplify algebraic fractions?
3. How does exponentiation of fractional bases affect results?
4. What is the importance of rationalizing denominators?
5. Can the difference of squares be applied in other types of problems?

**Tip:** Always remember to use the properties of exponents consistently when simplifying complex expressions with multiple variables and powers.

1. Simplify: ((a^(-3) * b^(-2) * c) / (a^(-5) * b^(-7) * c^2))^(-2)
2. Given a = 16, b = 625, c = 27, find the value of a^(1/2) * b^(-3/4) * c^(-2/3)
3. Rationalize the expression: (2 - sqrt(5)) / (2 + sqrt(5))

This problem covers key algebraic concepts such as simplifying exponential expressions with negative powers, calculating fractional exponents, and rationalizing denominators. It includes the simplification of ((a^(-3) * b^(-2) * c) / (a^(-5) * b^(-7) * c^2))^(-2), as well as evaluating values like a^(1/2), b^(-3/4), and c^(-2/3), and rationalizing the expression (2 - sqrt(5)) / (2 + sqrt(5)). Suitable for Grades 9-11.

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