b) \( \left(\frac{3}{5} m n^{2}-\frac{4 m}{2}\right)\left(\frac{6}{5} m^{2} n^{2}+\frac{3}{2} n\right)=\frac{18 m^{\square} n^{4}}{25} \cdot \frac{27 m^{4} n^{\square}}{5}+\frac{9 m n^{1}}{10}+\frac{27 m n}{4} \) b) \( \left(-6 m n^{1} n^{3}-7 m^{\square}\right)\left(5 m^{\square}+2 n-2\right)=-30 m^{6} n^{2}-12 m^{2} n^{5}+12 m^{2} n^{3}-35 m^{7} \) \( -14 m^{2} n^{2}+74 m^{3} \) c) \( 0.5 m n\left(-2.5 m^{2}+1.5 m^{3} n-9\right)=-1,25 m^{0} n^{0}+0,95 m^{9} n^{\square}-45 m n \)

To solve the equations provided, we will analyze each part step by step. ### Part b The expression is: \[ \left(\frac{3}{5} m n^{2}-\frac{4 m}{2}\right)\left(\frac{6}{5} m^{2} n^{2}+\frac{3}{2} n\right) \] First, simplify the first factor: \[ \frac{3}{5} m n^{2} - \frac{4 m}{2} = \frac{3}{5} m n^{2} - 2 m = m\left(\frac{3}{5} n^{2} - 2\right) \] Now simplify the second factor: \[ \frac{6}{5} m^{2} n^{2} + \frac{3}{2} n \] Now we multiply the two factors: \[ m\left(\frac{3}{5} n^{2} - 2\right)\left(\frac{6}{5} m^{2} n^{2} + \frac{3}{2} n\right) \] Distributing: 1. \( m \cdot \frac{3}{5} n^{2} \cdot \frac{6}{5} m^{2} n^{2} = \frac{18}{25} m^{3} n^{4} \) 2. \( m \cdot \frac{3}{5} n^{2} \cdot \frac{3}{2} n = \frac{9}{10} m n^{3} \) 3. \( m \cdot (-2) \cdot \frac{6}{5} m^{2} n^{2} = -\frac{12}{5} m^{3} n^{2} \) 4. \( m \cdot (-2) \cdot \frac{3}{2} n = -3 m n \) Combining these results: \[ \frac{18}{25} m^{3} n^{4} + \frac{9}{10} m n^{3} - \frac{12}{5} m^{3} n^{2} - 3 m n \] Now, we need to match this with: \[ \frac{18 m^{\square} n^{4}}{25} \cdot \frac{27 m^{4} n^{\square}}{5}+\frac{9 m n^{1}}{10}+\frac{27 m n}{4} \] From the first term, we can see that \( m^{\square} = 3 \) and \( n^{\square} = 4 \). ### Part b (continued) The second part of b is: \[ \left(-6 m n^{1} n^{3}-7 m^{\square}\right)\left(5 m^{\square}+2 n-2\right) \] Simplifying the first factor: \[ -6 m n^{4} - 7 m^{\square} \] Now simplifying the second factor: \[ 5 m^{\square} + 2 n - 2 \] Now we multiply: 1. \( (-6 m n^{4}) \cdot (5 m^{\square}) = -30 m^{1+\square} n^{4} \) 2. \( (-6 m n^{4}) \cdot (2 n) = -12 m n^{5} \) 3. \( (-6 m n^{4}) \cdot (-2) = 12 m n^{4} \) 4. \( (-7 m^{\square}) \cdot (5 m^{\square}) = -35 m^{2\square} \) 5. \( (-7 m^{\square}) \cdot (2 n) = -14 m^{\square} n \) 6. \( (-7 m^{\square}) \cdot (-2) = 14 m^{\square} \) Combining these results: \[ -30 m^{1+\square} n^{4} - 12 m n^{5} + 12 m n^{4} - 35 m^{2\square} - 14 m^{\square} n + 14 m^{\square} \] Now we need to match this with: \[ -30 m^{6} n^{2}-12 m^{2} n^{5}+12 m^{2} n^{3}-35 m^{7} -14 m^{2} n^{2}+74 m^{3} \] From this, we can deduce that \( \square = 2 \). ### Part c