Valdez Whittaker
07/11/2024 · Middle School
Recuperacion \( \left(\frac{2 a b^{2}}{3}-\frac{6 a b^{2}}{7}+\frac{2 b}{3}\right) \div \frac{2 b^{2}}{3} \) \( \left(8 m^{4} n^{2}-70 m^{4} n^{4}-20 m^{5} n^{6}+12 m^{3} n^{8}\right) \div 2 \)
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Step-by-step Solution
Para resolver las expresiones dadas, vamos a simplificarlas paso a paso.
### Primera expresión:
\[
\left(\frac{2 a b^{2}}{3}-\frac{6 a b^{2}}{7}+\frac{2 b}{3}\right) \div \frac{2 b^{2}}{3}
\]
1. **Simplificamos el numerador**:
- Para sumar las fracciones, necesitamos un común denominador. El mínimo común múltiplo de 3 y 7 es 21.
- Reescribimos cada término con el denominador 21:
\[
\frac{2 a b^{2}}{3} = \frac{14 a b^{2}}{21}, \quad \frac{6 a b^{2}}{7} = \frac{18 a b^{2}}{21}, \quad \frac{2 b}{3} = \frac{14 b}{21}
\]
- Ahora, sumamos:
\[
\frac{14 a b^{2}}{21} - \frac{18 a b^{2}}{21} + \frac{14 b}{21} = \frac{(14 a b^{2} - 18 a b^{2} + 14 b)}{21} = \frac{-4 a b^{2} + 14 b}{21}
\]
2. **Dividimos por \(\frac{2 b^{2}}{3}\)**:
\[
\frac{-4 a b^{2} + 14 b}{21} \div \frac{2 b^{2}}{3} = \frac{-4 a b^{2} + 14 b}{21} \cdot \frac{3}{2 b^{2}} = \frac{3(-4 a b^{2} + 14 b)}{42 b^{2}} = \frac{-12 a b^{2} + 42 b}{42 b^{2}}
\]
- Simplificamos:
\[
= \frac{-12 a b^{2}}{42 b^{2}} + \frac{42 b}{42 b^{2}} = \frac{-12 a}{42} + \frac{42}{42 b} = -\frac{2 a}{7} + \frac{1}{b}
\]
### Segunda expresión:
\[
\left(8 m^{4} n^{2}-70 m^{4} n^{4}-20 m^{5} n^{6}+12 m^{3} n^{8}\right) \div 2
\]
1. **Dividimos cada término por 2**:
\[
\frac{8 m^{4} n^{2}}{2} - \frac{70 m^{4} n^{4}}{2} - \frac{20 m^{5} n^{6}}{2} + \frac{12 m^{3} n^{8}}{2}
\]
- Esto nos da:
\[
4 m^{4} n^{2} - 35 m^{4} n^{4} - 10 m^{5} n^{6} + 6 m^{3} n^{8}
\]
### Resumen de resultados:
1. La primera expresión simplificada es:
\[
-\frac{2 a}{7} + \frac{1}{b}
\]
2. La segunda expresión simplificada es:
\[
4 m^{4} n^{2} - 35 m^{4} n^{4} - 10 m^{5} n^{6} + 6 m^{3} n^{8}
\]
Quick Answer
1. La primera expresión simplificada es:
\[
-\frac{2 a}{7} + \frac{1}{b}
\]
2. La segunda expresión simplificada es:
\[
4 m^{4} n^{2} - 35 m^{4} n^{4} - 10 m^{5} n^{6} + 6 m^{3} n^{8}
\]
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