To simplify the expression:

\[
\frac{2^{-2} + 2^{-3}}{4^{-2} + 8^{-1}}
\]

we need to rewrite the terms with negative exponents as positive exponents.

### Step 1: Rewrite with Positive Exponents

1. \(2^{-2} = \frac{1}{2^2} = \frac{1}{4}\)
2. \(2^{-3} = \frac{1}{2^3} = \frac{1}{8}\)
3. \(4^{-2} = \frac{1}{4^2} = \frac{1}{16}\) (since \(4 = 2^2\), \(4^{-2} = (2^2)^{-2} = 2^{-4} = \frac{1}{16}\))
4. \(8^{-1} = \frac{1}{8} = \frac{1}{2^3}\)

Substitute these into the original expression:

\[
\frac{\frac{1}{4} + \frac{1}{8}}{\frac{1}{16} + \frac{1}{8}}
\]

### Step 2: Simplify the Numerator and Denominator

To add fractions, we need a common denominator:

1. Numerator: \(\frac{1}{4} + \frac{1}{8} = \frac{2}{8} + \frac{1}{8} = \frac{3}{8}\)
2. Denominator: \(\frac{1}{16} + \frac{1}{8} = \frac{1}{16} + \frac{2}{16} = \frac{3}{16}\)

### Step 3: Divide the Fractions

Now we have:

\[
\frac{\frac{3}{8}}{\frac{3}{16}}
\]

To divide fractions, multiply by the reciprocal:

\[
\frac{3}{8} \times \frac{16}{3} = \frac{3 \times 16}{8 \times 3} = \frac{48}{24} = 2
\]

### Final Answer

\[
\frac{2^{-2} + 2^{-3}}{4^{-2} + 8^{-1}} = 2
\]

Would you like a breakdown of each step or have any questions? 

Here are 5 related questions to consider:
1. How do you simplify expressions with negative exponents?
2. What are the properties of exponents when multiplying or dividing powers with the same base?
3. How do you convert between different bases, such as converting \(4\) and \(8\) to powers of \(2\)?
4. What are the rules for adding and subtracting fractions?
5. How do you simplify complex fractions by multiplying by the reciprocal?

**Tip:** Always remember to convert negative exponents into positive exponents to make simplification easier.

Learn how to simplify the expression \frac{2^{-2} + 2^{-3}}{4^{-2} + 8^{-1}} step by step. This problem covers key algebraic concepts such as simplifying exponents, adding fractions, and dividing complex fractions. Suitable for students in Grades 8-10.

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