Question

1. $\frac{11^{-7} cdot 5^{9}}{6^{9}}$
2. $\frac{3^{5} cdot (-4)^{-10}}{7^{9} cdot 21^{-4}}$

Explanation:

Step1: Rewrite negative exponents

For problem 7: $11^{-7} = \frac{1}{11^7}$, so $\frac{11^{-7} \cdot 5^9}{6^9} = \frac{5^9}{11^7 \cdot 6^9}$
For problem 10: $(-4)^{-10} = \frac{1}{(-4)^{10}}$, $21^{-4} = \frac{1}{21^4}$, so $\frac{3^5 \cdot (-4)^{-10}}{7^9 \cdot 21^{-4}} = \frac{3^5 \cdot 21^4}{7^9 \cdot (-4)^{10}}$

Step2: Simplify $21^4$ (problem 10)

$21 = 3 \cdot 7$, so $21^4 = (3 \cdot 7)^4 = 3^4 \cdot 7^4$
Substitute: $\frac{3^5 \cdot 3^4 \cdot 7^4}{7^9 \cdot 4^{10}}$ (since $(-4)^{10}=4^{10}$)

Step3: Combine like bases (problem10)

Add exponents for $3$: $3^{5+4}=3^9$
Subtract exponents for $7$: $7^{4-9}=7^{-5}=\frac{1}{7^5}$
Result: $\frac{3^9}{7^5 \cdot 4^{10}}$

Answer:

Problem 7: $\frac{5^9}{11^7 \cdot 6^9}$
Problem 10: $\frac{3^9}{7^5 \cdot 4^{10}}$