Simplify rational expressions by multiplying or dividing — a math worksheet with seven problems.

Problem: Simplify the rational expressions by multiplying or dividing.

We will solve each problem step by step.

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#### 1) Simplify: \( \frac{4}{3} \times \frac{5}{7} \)

- Step 1: Multiply the numerators and the denominators.
\[
\frac{4}{3} \times \frac{5}{7} = \frac{4 \times 5}{3 \times 7} = \frac{20}{21}
\]

- Step 2: Check if the fraction can be simplified. The greatest common divisor (GCD) of 20 and 21 is 1, so the fraction is already in its simplest form.

Answer:
\[
\boxed{\frac{20}{21}}
\]

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#### 2) Simplify: \( \frac{5}{8} \times \frac{4}{15} \)

- Step 1: Multiply the numerators and the denominators.
\[
\frac{5}{8} \times \frac{4}{15} = \frac{5 \times 4}{8 \times 15} = \frac{20}{120}
\]

- Step 2: Simplify the fraction by finding the GCD of 20 and 120. The GCD is 20.
\[
\frac{20}{120} = \frac{20 \div 20}{120 \div 20} = \frac{1}{6}
\]

Answer:
\[
\boxed{\frac{1}{6}}
\]

---

#### 3) Simplify: \( \frac{7}{22} \times \frac{33}{35} \)

- Step 1: Multiply the numerators and the denominators.
\[
\frac{7}{22} \times \frac{33}{35} = \frac{7 \times 33}{22 \times 35} = \frac{231}{770}
\]

- Step 2: Simplify the fraction by finding the GCD of 231 and 770. The GCD is 77.
\[
\frac{231}{770} = \frac{231 \div 77}{770 \div 77} = \frac{3}{10}
\]

Answer:
\[
\boxed{\frac{3}{10}}
\]

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#### 4) Simplify: \( \frac{40x^2y^2}{24xy^4} \div \frac{27xy}{8x^2y^3} \)

- Step 1: Rewrite the division as multiplication by the reciprocal.
\[
\frac{40x^2y^2}{24xy^4} \div \frac{27xy}{8x^2y^3} = \frac{40x^2y^2}{24xy^4} \times \frac{8x^2y^3}{27xy}
\]

- Step 2: Multiply the numerators and the denominators.
\[
\frac{40x^2y^2 \cdot 8x^2y^3}{24xy^4 \cdot 27xy} = \frac{320x^4y^5}{648x^2y^5}
\]

- Step 3: Simplify the coefficients and the variables.
- Simplify the coefficients: \( \frac{320}{648} \). The GCD of 320 and 648 is 16.
\[
\frac{320}{648} = \frac{320 \div 16}{648 \div 16} = \frac{20}{41}
\]
- Simplify the variables:
\[
\frac{x^4y^5}{x^2y^5} = x^{4-2}y^{5-5} = x^2y^0 = x^2
\]

- Step 4: Combine the results.
\[
\frac{20}{41} \cdot x^2 = \frac{20x^2}{41}
\]

Answer:
\[
\boxed{\frac{20x^2}{41}}
\]

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#### 5) Simplify: \( \frac{x^2 - y^2}{x^2 - y^2} \div \frac{x + y}{x} \)

- Step 1: Rewrite the division as multiplication by the reciprocal.
\[
\frac{x^2 - y^2}{x^2 - y^2} \div \frac{x + y}{x} = \frac{x^2 - y^2}{x^2 - y^2} \times \frac{x}{x + y}
\]

- Step 2: Factor the numerator and denominator where possible.
- \( x^2 - y^2 \) is a difference of squares: \( x^2 - y^2 = (x + y)(x - y) \).
\[
\frac{(x + y)(x - y)}{(x + y)(x - y)} \times \frac{x}{x + y}
\]

- Step 3: Simplify the fractions.
- The \( (x + y)(x - y) \) terms cancel out in the first fraction.
\[
\frac{(x + y)(x - y)}{(x + y)(x - y)} = 1
\]
- So the expression becomes:
\[
1 \times \frac{x}{x + y} = \frac{x}{x + y}
\]

Answer:
\[
\boxed{\frac{x}{x + y}}
\]

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#### 6) Simplify: \( \frac{x^2 - 6x + 9}{x^2 - 7x + 12} \times \frac{x^2 - 4x^2 + 9x - 36}{x^4 - 81} \)

- Step 1: Simplify each polynomial in the expression.
- Factor \( x^2 - 6x + 9 \):
\[
x^2 - 6x + 9 = (x - 3)^2
\]
- Factor \( x^2 - 7x + 12 \):
\[
x^2 - 7x + 12 = (x - 3)(x - 4)
\]
- Factor \( x^2 - 4x^2 + 9x - 36 \):
\[
x^2 - 4x^2 + 9x - 36 = -3x^2 + 9x - 36 = -3(x^2 - 3x + 12)
\]
(Note: This polynomial does not factor further over the integers.)
- Factor \( x^4 - 81 \):
\[
x^4 - 81 = (x^2)^2 - 9^2 = (x^2 - 9)(x^2 + 9) = (x - 3)(x + 3)(x^2 + 9)
\]

- Step 2: Substitute the factored forms into the expression.
\[
\frac{(x - 3)^2}{(x - 3)(x - 4)} \times \frac{-3(x^2 - 3x + 12)}{(x - 3)(x + 3)(x^2 + 9)}
\]

- Step 3: Simplify the expression.
- Cancel the common factors:
\[
\frac{(x - 3)^2}{(x - 3)(x - 4)} = \frac{x - 3}{x - 4} \quad \text{(cancel one } x - 3 \text{)}
\]
\[
\frac{-3(x^2 - 3x + 12)}{(x - 3)(x + 3)(x^2 + 9)}
\]
- Combine the simplified fractions:
\[
\frac{x - 3}{x - 4} \times \frac{-3(x^2 - 3x + 12)}{(x - 3)(x + 3)(x^2 + 9)}
\]
- Cancel the \( x - 3 \) term:
\[
\frac{1}{x - 4} \times \frac{-3(x^2 - 3x + 12)}{(x + 3)(x^2 + 9)}
\]
- Multiply the remaining terms:
\[
\frac{-3(x^2 - 3x + 12)}{(x - 4)(x + 3)(x^2 + 9)}
\]

Answer:
\[
\boxed{\frac{-3(x^2 - 3x + 12)}{(x - 4)(x + 3)(x^2 + 9)}}
\]

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Final Answers:

1. \( \boxed{\frac{20}{21}} \)
2. \( \boxed{\frac{1}{6}} \)
3. \( \boxed{\frac{3}{10}} \)
4. \( \boxed{\frac{20x^2}{41}} \)
5. \( \boxed{\frac{x}{x + y}} \)
6. \( \boxed{\frac{-3(x^2 - 3x + 12)}{(x - 4)(x + 3)(x^2 + 9)}} \)