d) \( \left[\left(\frac{1}{3} \times \frac{3}{2}+\frac{3}{4}\right)-\left(\frac{17}{2}-\frac{6}{5} \div \frac{3}{20}\right)\right] \div \frac{9}{4} \)

To solve \( \left[\left(\frac{1}{3} \times \frac{3}{2}+\frac{3}{4}\right)-\left(\frac{17}{2}-\frac{6}{5} \div \frac{3}{20}\right)\right] \div \frac{9}{4} \), we’ll simplify each part step by step. First, we calculate the expression inside the brackets: 1. **Calculate \( \frac{1}{3} \times \frac{3}{2} \):** \[ \frac{1}{3} \times \frac{3}{2} = \frac{1 \times 3}{3 \times 2} = \frac{3}{6} = \frac{1}{2}. \] 2. **Add \( \frac{3}{4} \) to \( \frac{1}{2} \):** To add these fractions, we need a common denominator. The least common denominator for \( 2 \) and \( 4 \) is \( 4 \). \[ \frac{1}{2} = \frac{2}{4}, \quad \text{so} \quad \frac{2}{4} + \frac{3}{4} = \frac{5}{4}. \] 3. **Calculate \( \frac{6}{5} \div \frac{3}{20} \):** Dividing by a fraction is the same as multiplying by its reciprocal. \[ \frac{6}{5} \div \frac{3}{20} = \frac{6}{5} \times \frac{20}{3} = \frac{120}{15} = 8. \] 4. **Subtract \( 8 \) from \( \frac{17}{2} \):** Again, we need a common denominator. The least common denominator for \( 2 \) and \( 1 \) is \( 2 \). \[ 8 = \frac{16}{2}, \quad \text{therefore} \quad \frac{17}{2} - \frac{16}{2} = \frac{1}{2}. \] Now we put it all back into the main expression: 5. **Subtract \( \frac{1}{2} \) from \( \frac{5}{4} \):** We again find a common denominator (which is \( 4 \)). \[ \frac{1}{2} = \frac{2}{4}, \quad \text{thus} \quad \frac{5}{4} - \frac{2}{4} = \frac{3}{4}. \] Finally, we have: \[ \left[\frac{3}{4}\right] \div \frac{9}{4}. \] 6. **Divide \( \frac{3}{4} \) by \( \frac{9}{4} \):** \[ \frac{3}{4} \div \frac{9}{4} = \frac{3}{4} \times \frac{4}{9} = \frac{3 \cdot 4}{4 \cdot 9} = \frac{3}{9} = \frac{1}{3}. \] Thus, the final result is: \[ \boxed{\frac{1}{3}}. \]