Whittaker Huang
06/02/2024 · Senior High School
a) \( \left(\frac{3}{5} m n^{2}-\frac{9 m}{2}\right)\left(\frac{6}{5} m^{3} n^{2}+\frac{3}{2} n\right)=\frac{18 m^{\square} n^{4}}{25}-\frac{27 m^{4} n^{\square}}{5}+\frac{9 m n^{\square}}{10}+2 \) b) \( \left(-6 m n^{\square} n^{3}-7 m^{\square}\right)\left(5 m^{\square}+2 n-2\right)=-30 m^{6} n^{3}-12 m^{2} n^{5}+12 m^{2} n^{3}-35 m \) \( -14 m^{3} n^{2}+14 m^{3} \) c) \( 0.5 m n\left(-2.5 m^{2}+1,5 m^{3} n-9\right)=-1,25 m^{\square} n^{\square}+0.75 m^{\square} n^{\square}-4.5 m \)
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Step-by-step Solution
To solve the problems, we need to expand the expressions and match the coefficients to find the missing exponents represented by the squares.
### Part a
We need to expand:
\[
\left(\frac{3}{5} m n^{2}-\frac{9 m}{2}\right)\left(\frac{6}{5} m^{3} n^{2}+\frac{3}{2} n\right)
\]
1. **First Term**:
\[
\frac{3}{5} m n^{2} \cdot \frac{6}{5} m^{3} n^{2} = \frac{18}{25} m^{4} n^{4}
\]
2. **Second Term**:
\[
\frac{3}{5} m n^{2} \cdot \frac{3}{2} n = \frac{9}{10} m n^{3}
\]
3. **Third Term**:
\[
-\frac{9 m}{2} \cdot \frac{6}{5} m^{3} n^{2} = -\frac{54}{10} m^{4} n^{2} = -\frac{27}{5} m^{4} n^{2}
\]
4. **Fourth Term**:
\[
-\frac{9 m}{2} \cdot \frac{3}{2} n = -\frac{27}{4} m n
\]
Combining these results:
\[
\frac{18}{25} m^{4} n^{4} + \frac{9}{10} m n^{3} - \frac{27}{5} m^{4} n^{2} - \frac{27}{4} m n
\]
Now, we can match the coefficients with the given expression:
\[
\frac{18 m^{\square} n^{4}}{25} - \frac{27 m^{4} n^{\square}}{5} + \frac{9 m n^{\square}}{10} + 2
\]
From the expansion:
- \( m^{\square} = 4 \) (since \( m^{4} \))
- \( n^{\square} = 2 \) (since \( n^{2} \))
- \( n^{\square} = 3 \) (since \( n^{3} \))
### Part b
We need to expand:
\[
\left(-6 m n^{\square} n^{3}-7 m^{\square}\right)\left(5 m^{\square}+2 n-2\right)
\]
1. **First Term**:
\[
-6 m n^{\square} n^{3} \cdot 5 m^{\square} = -30 m^{\square + 1} n^{\square + 3}
\]
2. **Second Term**:
\[
-6 m n^{\square} n^{3} \cdot 2 n = -12 m n^{\square + 4}
\]
3. **Third Term**:
\[
-6 m n^{\square} n^{3} \cdot -2 = 12 m n^{\square + 3}
\]
4. **Fourth Term**:
\[
-7 m^{\square} \cdot 5 m^{\square} = -35 m^{2\square}
\]
5. **Fifth Term**:
\[
-7 m^{\square} \cdot 2 n = -14 m^{\square} n
\]
6. **Sixth Term**:
\[
-7 m^{\square} \cdot -2 = 14 m^{\square}
\]
Combining these results:
\[
-30 m^{\square + 1} n^{\square + 3} - 12 m n^{\square + 4} + 12 m n^{\square + 3} - 35 m^{2\square} - 14 m^{\square} n + 14 m^{\square}
\]
Now, we can match the coefficients with the given expression:
\[
-30 m^{6} n^{3} - 12 m^{2} n^{5} + 12 m^{2} n^{3} - 35 m - 14 m^{3} n^{2} + 14 m^{3}
\]
From the expansion:
- \( n^{\square} = 3 \) (since \(
Quick Answer
a) \( m^{\square} = 4, n^{\square} = 2, n^{\square} = 3 \)
b) \( n^{\square} = 3, m^{\square} = 2, m^{\square} = 3 \)
c) \( m^{\square} = 2, n^{\square} = 2, m^{\square} = 2, n^{\square} = 2 \)
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