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Let's simplify the given mathematical expressions, which are written in Indonesian. The task is to simplify the expressions to positive integer exponents.

### a. Simplify:
\[
625 \left( \frac{a^3 b^2 c}{5a^4 b^5} \right)^2
\]

1. Simplify the fraction inside the parentheses:
   \[
   \frac{a^3 b^2 c}{5 a^4 b^5} = \frac{a^{3-4} b^{2-5} c}{5} = \frac{a^{-1} b^{-3} c}{5}
   \]

2. Square the expression:
   \[
   \left( \frac{a^{-1} b^{-3} c}{5} \right)^2 = \frac{a^{-2} b^{-6} c^2}{25}
   \]

3. Multiply by 625:
   \[
   625 \times \frac{a^{-2} b^{-6} c^2}{25} = 25 a^{-2} b^{-6} c^2
   \]

So the simplified result for part (a) is:
\[
25 a^{-2} b^{-6} c^2
\]

---

### b. Simplify:
\[
\left( \frac{3x^2 y^3}{2ab^4} \right)^6 \left( \frac{16 a^3 b^6}{27 x y^3} \right)^2
\]

1. Simplify the first term:
   \[
   \left( \frac{3x^2 y^3}{2ab^4} \right)^6 = \frac{3^6 x^{12} y^{18}}{2^6 a^6 b^{24}} = \frac{729 x^{12} y^{18}}{64 a^6 b^{24}}
   \]

2. Simplify the second term:
   \[
   \left( \frac{16 a^3 b^6}{27 x y^3} \right)^2 = \frac{16^2 a^6 b^{12}}{27^2 x^2 y^6} = \frac{256 a^6 b^{12}}{729 x^2 y^6}
   \]

3. Multiply the two terms together:
   \[
   \frac{729 x^{12} y^{18}}{64 a^6 b^{24}} \times \frac{256 a^6 b^{12}}{729 x^2 y^6} = \frac{729 \times 256 x^{12-2} y^{18-6} a^6 b^{12}}{64 \times 729 a^6 b^{24}} = \frac{256 x^{10} y^{12}}{64 b^{12}}
   \]

4. Simplify:
   \[
   \frac{256}{64} = 4
   \]

So the simplified result for part (b) is:
\[
4 x^{10} y^{12} b^{-12}
\]

---

### c. Simplify:
\[
\frac{a^{-2} + b^{-1}}{a^{-1} + b^{-2}}
\]

1. Use properties of negative exponents:
   \[
   \frac{\frac{1}{a^2} + \frac{1}{b}}{\frac{1}{a} + \frac{1}{b^2}} = \frac{\frac{b + a^2}{a^2b}}{\frac{b^2 + a}{ab^2}} = \frac{b + a^2}{a^2 b} \times \frac{ab^2}{b^2 + a} = \frac{(b + a^2) a b^2}{a^2 b (b^2 + a)}
   \]

2. Simplify the expression:
   \[
   = \frac{(b + a^2) b}{a(b^2 + a)}
   \]

So the simplified result for part (c) is:
\[
\frac{(b + a^2) b}{a (b^2 + a)}
\]

---

### d. Simplify:
\[
\frac{2^{-3} + 3^{-2}}{2^{-2} + 3^{-3}}
\]

1. Calculate each part using exponent properties:
   \[
   2^{-3} = \frac{1}{8}, \quad 3^{-2} = \frac{1}{9}, \quad 2^{-2} = \frac{1}{4}, \quad 3^{-3} = \frac{1}{27}
   \]

2. Simplify the numerator:
   \[
   \frac{1}{8} + \frac{1}{9} = \frac{9 + 8}{72} = \frac{17}{72}
   \]

3. Simplify the denominator:
   \[
   \frac{1}{4} + \frac{1}{27} = \frac{27 + 4}{108} = \frac{31}{108}
   \]

4. Divide the fractions:
   \[
   \frac{\frac{17}{72}}{\frac{31}{108}} = \frac{17}{72} \times \frac{108}{31} = \frac{17 \times 108}{72 \times 31} = \frac{17 \times 3}{31} = \frac{51}{31}
   \]

So the simplified result for part (d) is:
\[
\frac{51}{31}
\]

---

Would you like further details or have any questions?

Here are 5 related questions:
1. How do you simplify expressions with negative exponents?
2. Can you explain the properties of exponents used in part (a)?
3. How do we handle multiplication of fractional expressions in part (b)?
4. What is the general strategy to simplify complex fractions like in part (c)?
5. How do you convert negative exponents to positive exponents?

**Tip**: Always break down the simplification step-by-step, especially when dealing with multiple exponents.

Sederhanakan bentuk-bentuk berikut dalam bentuk pangkat bulat positif.

Step-by-step simplification of algebraic expressions involving exponents, including negative exponents and complex fractions. Suitable for students in Grades 10-12.

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