1. Efectúa las siguientes operaciones. a. \( \frac{5}{2}+\left(-\frac{1}{2}\right) \) b. \( (\sqrt{2}+\sqrt{5})+(-\sqrt{5}) \quad \frac{1}{2}+\frac{3}{4} \cdot \frac{6}{15} \) c. \( \frac{\sqrt{2}}{2}-\sqrt{2}-\left(4 \sqrt{2}-\frac{\sqrt{2}}{3}\right) \) \( \left(\frac{4}{7}+\frac{1}{2}\right) \div \frac{5}{28} \)
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Claro, vamos a resolver cada una de las operaciones paso a paso. a. \( \frac{5}{2}+\left(-\frac{1}{2}\right) = \frac{5}{2} - \frac{1}{2} = \frac{5-1}{2} = \frac{4}{2} = 2 \) b. \( (\sqrt{2}+\sqrt{5})+(-\sqrt{5}) = \sqrt{2} + \sqrt{5} - \sqrt{5} = \sqrt{2} \) Luego, calculamos \( \frac{1}{2}+\frac{3}{4} \cdot \frac{6}{15} \): Primero multiplicamos \( \frac{3}{4} \cdot \frac{6}{15} = \frac{18}{60} = \frac{3}{10} \). Ahora, sumamos \( \frac{1}{2} + \frac{3}{10} = \frac{5}{10} + \frac{3}{10} = \frac{8}{10} = \frac{4}{5} \). c. \( \frac{\sqrt{2}}{2}-\sqrt{2}-\left(4 \sqrt{2}-\frac{\sqrt{2}}{3}\right) \): Primero transformamos \( \sqrt{2} \) a fracción: \( \sqrt{2} = \frac{\sqrt{2}}{1} \). Entonces: \[ = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{1} - 4\sqrt{2} + \frac{\sqrt{2}}{3} \] Hacemos un común denominador que es 6: \[ = \frac{3\sqrt{2}}{6} - \frac{6\sqrt{2}}{6} - \frac{24\sqrt{2}}{6} + \frac{2\sqrt{2}}{6} = \frac{3\sqrt{2} - 6\sqrt{2} - 24\sqrt{2} + 2\sqrt{2}}{6} = \frac{-25\sqrt{2}}{6} \] Ahora calculamos \( \left(\frac{4}{7}+\frac{1}{2}\right) \div \frac{5}{28} \): Primero, sumamos \( \frac{4}{7} + \frac{1}{2} = \frac{8}{14} + \frac{7}{14} = \frac{15}{14} \). Luego, hacemos la división: \[ \frac{15}{14} \div \frac{5}{28} = \frac{15}{14} \cdot \frac{28}{5} = \frac{15 \cdot 28}{14 \cdot 5} = \frac{420}{70} = 6. \] Entonces los resultados finales son: a. \( 2 \) b. \( \sqrt{2} \) y \( \frac{4}{5} \) c. \( \frac{-25\sqrt{2}}{6} \) y \( 6 \)
