18. \( \frac{b^{2}+4 b+4}{2 b^{2}-8} \cdot \frac{3 b-6}{4 b} \)

Let's simplify the given expression step by step: 1. **Factor all parts of the expression** where possible: - The numerator \( b^2 + 4b + 4 \) can be factored as \( (b + 2)^2 \). - The denominator \( 2b^2 - 8 \) can be factored as \( 2(b^2 - 4) = 2(b - 2)(b + 2) \). - The numerator \( 3b - 6 \) can be factored as \( 3(b - 2) \). - The denominator \( 4b \) remains as is. Now, we can rewrite the entire expression with the factors: \[ \frac{(b + 2)^2}{2(b - 2)(b + 2)} \cdot \frac{3(b - 2)}{4b} \] 2. **Cancel common factors**: - The \( (b + 2) \) in the numerator of the first fraction can cancel with the \( (b + 2) \) in the denominator. - The \( (b - 2) \) in the numerator of the second fraction can cancel with the \( (b - 2) \) in the denominator from the first fraction. After canceling, we get: \[ \frac{(b + 2) \cdot 3}{2 \cdot 4b} = \frac{3(b + 2)}{8b} \] 3. **Final expression**: Thus, the simplified expression is: \[ \frac{3(b + 2)}{8b} \] This is the final result of simplifying the initial expression.