Ejercicios de Extracción de factor común

Desarrollo:

1. ${\displaystyle \frac{1}{5}}\cdot {\displaystyle \frac{2}{3}}+{\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{1}{5}}-{\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{1}{5}}={\displaystyle \frac{1}{5}}\cdot ({\displaystyle \frac{2}{3}}+{\displaystyle \frac{1}{3}}-{\displaystyle \frac{1}{3}})=$ $={\displaystyle \frac{1}{5}}\cdot {\displaystyle \frac{2}{3}}={\displaystyle \frac{2}{15}}$

2. ${\displaystyle \frac{3}{5}}\cdot {\displaystyle \frac{a}{b}}-{\displaystyle \frac{2}{b}}\cdot {\displaystyle \frac{a}{3}}+{\displaystyle \frac{a}{b}}\cdot {\displaystyle \frac{1}{3}}={\displaystyle \frac{3}{5}}\cdot {\displaystyle \frac{a}{b}}-{\displaystyle \frac{a}{b}}\cdot {\displaystyle \frac{2}{3}}+{\displaystyle \frac{a}{b}}\cdot {\displaystyle \frac{1}{3}}=$ $={\displaystyle \frac{3}{5}}\cdot {\displaystyle \frac{a}{b}}+{\displaystyle \frac{a}{b}}\cdot (-{\displaystyle \frac{2}{3}})+{\displaystyle \frac{a}{b}}\cdot {\displaystyle \frac{1}{3}}={\displaystyle \frac{a}{b}}\cdot ({\displaystyle \frac{3}{5}}-{\displaystyle \frac{2}{3}}+{\displaystyle \frac{1}{3}})=$ ${\displaystyle \frac{a}{b}}\cdot ({\displaystyle \frac{9}{15}}-{\displaystyle \frac{10}{15}}+{\displaystyle \frac{5}{15}})={\displaystyle \frac{a}{b}}\cdot ({\displaystyle \frac{9-10+5}{15}})={\displaystyle \frac{a}{b}}\cdot {\displaystyle \frac{4}{15}}={\displaystyle \frac{4a}{15b}}$

Solución:

1. ${\displaystyle \frac{2}{15}}$
2. ${\displaystyle \frac{4a}{15b}}$

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