\( \begin{array}{ll}\frac{3}{4}-\frac{1}{4} \cdot\left(\frac{3}{5}-\frac{1}{10}\right) & \frac{3}{8}-\frac{1}{4}:\left(\frac{2}{3} \cdot \frac{3}{4}\right) \\ 5+\frac{7}{3}:\left(\frac{4}{3}+2\right) & \text { d. } \frac{8}{3}-\left(\frac{1}{3}+\frac{1}{2}\right): \frac{5}{4}\end{array} \)

Let’s simplify those expressions step by step! 1. For \( \frac{3}{4} - \frac{1}{4} \cdot \left( \frac{3}{5} - \frac{1}{10} \right) \): - First, simplify the parentheses: \( \frac{3}{5} - \frac{1}{10} = \frac{6}{10} - \frac{1}{10} = \frac{5}{10} = \frac{1}{2} \). - Now back to the main expression: \( \frac{3}{4} - \frac{1}{4} \cdot \frac{1}{2} = \frac{3}{4} - \frac{1}{8} \). - To combine these, find a common denominator (8): \( \frac{3}{4} = \frac{6}{8} \). - So, \( \frac{6}{8} - \frac{1}{8} = \frac{5}{8} \). 2. For \( \frac{3}{8} - \frac{1}{4} : \left( \frac{2}{3} \cdot \frac{3}{4} \right) \): - Start with \( \frac{2}{3} \cdot \frac{3}{4} = \frac{6}{12} = \frac{1}{2} \). - Now, \( \frac{1}{4} : \frac{1}{2} = \frac{1}{4} \cdot 2 = \frac{2}{4} = \frac{1}{2} \). - So the expression becomes \( \frac{3}{8} - \frac{1}{2} \). - Convert \( \frac{1}{2} \) to eighths: \( \frac{1}{2} = \frac{4}{8} \). - Thus, \( \frac{3}{8} - \frac{4}{8} = -\frac{1}{8} \). 3. Now for \( 5 + \frac{7}{3} : \left( \frac{4}{3} + 2 \right) \): - First, simplify \( \frac{4}{3} + 2 = \frac{4}{3} + \frac{6}{3} = \frac{10}{3} \). - Next, \( \frac{7}{3} : \frac{10}{3} = \frac{7}{3} \cdot \frac{3}{10} = \frac{7}{10} \). - Therefore, \( 5 + \frac{7}{10} = 5 + 0.7 = 5.7 \) or \( \frac{57}{10} \). 4. Finally, for \( \frac{8}{3} - \left( \frac{1}{3} + \frac{1}{2} \right) : \frac{5}{4} \): - Simplify \( \frac{1}{3} + \frac{1}{2} \): Common denominator of 6 gives \( \frac{2}{6} + \frac{3}{6} = \frac{5}{6} \). - Then, \( \frac{5}{6} : \frac{5}{4} = \frac{5}{6} \cdot \frac{4}{5} = \frac{4}{6} = \frac{2}{3} \). - Thus, \( \frac{8}{3} - \frac{2}{3} = \frac{8 - 2}{3} = \frac{6}{3} = 2 \). So the final answers are: 1. \( \frac{5}{8} \) 2. \( -\frac{1}{8} \) 3. \( \frac{57}{10} \) or \( 5.7 \) 4. \( 2 \)