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Solve the following multiplications:

[tex]\[
\begin{array}{cc}
1(-3a) & \left(a^2\right) \\
\left(3x^2\right) & \left(-x^3 n\right) \\
\left(-m^2 n\right) & \left(-3m^2\right) \\
\left(-5a^3 y^2\right) & \left(a y^2\right) \\
\left(-2ab\right) & \left(-3a^2 b^3\right) \\
\left(\frac{1}{2} x^3\right) & \left(-\frac{2}{3} a^2 x\right) \\
\left(\frac{2}{3} a^5\right) & \left(\frac{2}{4} a^2 b^4\right) \\
8\left(-\frac{3}{5} m^2\right) & \left(-5 a^2 m\right) \\
q\left(-\frac{1}{2} x^2 y\right) & \left(-\frac{3}{5} x y^2\right) \\
0\left(-\frac{10}{3} x^3\right) & \left(-\frac{3}{4} x^2 y\right)
\end{array}
\][/tex]


Sagot :

¡Claro! Resolveré cada una de las multiplicaciones paso a paso:

### 1. [tex]\(1(-3a) \left(a^2\right)\)[/tex]
[tex]\[ 1 \cdot (-3a) \cdot a^2 = -3a \cdot a^2 = -3a^3 \][/tex]

### 2. [tex]\(\left(3x^2\right)\left(-x^3n\right)\)[/tex]
[tex]\[ 3x^2 \cdot (-x^3n) = 3 \cdot (-1) \cdot x^{2+3} \cdot n = -3x^5n \][/tex]

### 3. [tex]\(\left(-m^2n\right)\left(-3m^2\right)\)[/tex]
[tex]\[ (-m^2n) \cdot (-3m^2) = (-1) \cdot (-3) \cdot m^{2+2} \cdot n = 3m^4n \][/tex]

### 4. [tex]\((-5a^3y^2)\left(ay^2\right)\)[/tex]
[tex]\[ -5a^3y^2 \cdot ay^2 = -5 \cdot a^{3+1} \cdot y^{2+2} = -5a^4y^4 \][/tex]

### 5. [tex]\((-2ab)\left(-3a^2b^3\right)\)[/tex]
[tex]\[ (-2ab) \cdot (-3a^2b^3) = (-2) \cdot (-3) \cdot a^{1+2} \cdot b^{1+3} = 6a^3b^4 \][/tex]

### 6. [tex]\(\left(\frac{1}{2}x^3\right)\left(-\frac{2}{3}a^2x\right)\)[/tex]
[tex]\[ \left(\frac{1}{2}x^3\right) \cdot \left(-\frac{2}{3}a^2x\right) = \frac{1}{2} \cdot -\frac{2}{3} \cdot a^2 \cdot x^{3+1}\][/tex]
[tex]\[ = -\frac{2}{6}a^2x^4 = -\frac{1}{3}a^2x^4 \][/tex]

### 7. [tex]\(\left(\frac{2}{3}a^5\right)\left(\frac{2}{4}a^2b^4\right)\)[/tex]
[tex]\[ \left(\frac{2}{3}a^5\right) \cdot \left(\frac{2}{4}a^2b^4\right) = \frac{2}{3} \cdot \frac{1}{2} \cdot a^{5+2} \cdot b^4 = \frac{2 \cdot 1}{3 \cdot 2} a^7b^4\][/tex]
[tex]\[ = \frac{1}{3}a^7b^4 \][/tex]

### 8. [tex]\(8\left(-\frac{3}{5}m^2\right)\left(-5a^2m\right)\)[/tex]
[tex]\[ 8 \cdot \left(-\frac{3}{5}m^2\right) \cdot \left(-5a^2m\right) = 8 \cdot -\frac{3}{5} \cdot -5 \cdot a^2 \cdot m^{2+1} \][/tex]
[tex]\[ = 8 \cdot \frac{15}{5} \cdot a^2 m^3 = 8 \cdot 3 \cdot a^2 m^3 = 24a^2m^3 \][/tex]

### 9. [tex]\(q\left(-\frac{1}{2}x^2y\right)\left(-\frac{3}{5}xy^2\right)\)[/tex]
[tex]\[ q \cdot \left(-\frac{1}{2}x^2y\right) \cdot \left(-\frac{3}{5}xy^2\right) = q \cdot -\frac{1}{2} \cdot -\frac{3}{5} \cdot x^{2+1} \cdot y^{1+2} \][/tex]
[tex]\[ = q \cdot \frac{3}{10} \cdot x^3y^3 = \frac{3}{10}q x^3y^3 \][/tex]

### 10. [tex]\(0\left(-\frac{10}{3}x^3\right)\left(-\frac{3}{4}x^2y\right)\)[/tex]
[tex]\[ 0 \cdot \left(-\frac{10}{3}x^3\right) \cdot \left(-\frac{3}{4}x^2y\right) = 0 \][/tex]

En resumen, las soluciones son:

[tex]\[ \begin{array}{cc} -3a^3, & -3n x^5, \\ 3m^4 n, & -5a^4 y^4, \\ 6a^3b^4, & -\frac{1}{3}a^2x^4, \\ \frac{1}{3}a^7b^4, & 24a^2m^3, \\ 0.3qx^3y^3, & 0 \end{array} \][/tex]
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