1. **State the problem:** Simplify the expression $$\frac{(2x^{-3}y^{-2})^{-2} (5x^{-4}y^{2})^{2}}{(10x^{2}y^{-5})^{1}}$$. 2. **Recall the exponent rules:** - $(a^m)^n = a^{mn}$ - $a^{-m} = \frac{1}{a^m}$ - When multiplying like bases, add exponents: $a^m \cdot a^n = a^{m+n}$ - When dividing like bases, subtract exponents: $\frac{a^m}{a^n} = a^{m-n}$ 3. **Apply the power to each term inside the parentheses:** - $(2x^{-3}y^{-2})^{-2} = 2^{-2} x^{(-3)(-2)} y^{(-2)(-2)} = \frac{1}{2^2} x^{6} y^{4} = \frac{1}{4} x^{6} y^{4}$ - $(5x^{-4}y^{2})^{2} = 5^{2} x^{-8} y^{4} = 25 x^{-8} y^{4}$ - $(10x^{2}y^{-5})^{1} = 10 x^{2} y^{-5}$ 4. **Multiply the numerators:** $$\frac{1}{4} x^{6} y^{4} \times 25 x^{-8} y^{4} = \frac{25}{4} x^{6 + (-8)} y^{4 + 4} = \frac{25}{4} x^{-2} y^{8}$$ 5. **Divide by the denominator:** $$\frac{\frac{25}{4} x^{-2} y^{8}}{10 x^{2} y^{-5}} = \frac{25}{4} \times \frac{1}{10} x^{-2 - 2} y^{8 - (-5)} = \frac{25}{40} x^{-4} y^{13} = \frac{5}{8} x^{-4} y^{13}$$ 6. **Rewrite negative exponents as positive:** $$\frac{5}{8} \frac{y^{13}}{x^{4}}$$ **Final answer:** $$\frac{5 y^{13}}{8 x^{4}}$$