Rodriquez Cervantes
05/02/2023 · Senior High School

29. Escriban en lenguaje coloquial y resuelvan. a. \( \frac{3}{7}+\sqrt{\frac{1}{9}} \) b. \( 2 \cdot\left(\frac{4}{3}-\frac{2}{6}\right)^{2} \) c. \( \left(\frac{2}{6}\right)^{2}+\sqrt[3]{\frac{1}{27}} \) d. \( \frac{2}{5} \cdot \frac{10}{8}+\frac{1}{4} \) e. \( \sqrt{2 \cdot\left(\frac{22}{16}+\frac{133}{8}\right)} \)

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¡Claro! Vamos a resolver cada uno de los ejercicios paso a paso y en un lenguaje coloquial. ### a. \( \frac{3}{7}+\sqrt{\frac{1}{9}} \) Primero, resolvemos la raíz cuadrada: \[ \sqrt{\frac{1}{9}} = \frac{1}{3} \] Ahora sumamos: \[ \frac{3}{7} + \frac{1}{3} \] Para sumar estas fracciones, necesitamos un común denominador. El mínimo común múltiplo de 7 y 3 es 21. Entonces, convertimos las fracciones: \[ \frac{3}{7} = \frac{3 \cdot 3}{7 \cdot 3} = \frac{9}{21} \] \[ \frac{1}{3} = \frac{1 \cdot 7}{3 \cdot 7} = \frac{7}{21} \] Ahora sumamos: \[ \frac{9}{21} + \frac{7}{21} = \frac{16}{21} \] **Resultado:** \( \frac{16}{21} \) --- ### b. \( 2 \cdot\left(\frac{4}{3}-\frac{2}{6}\right)^{2} \) Primero, resolvemos la resta dentro del paréntesis: \[ \frac{2}{6} = \frac{1}{3} \] Entonces: \[ \frac{4}{3} - \frac{1}{3} = \frac{3}{3} = 1 \] Ahora elevamos al cuadrado: \[ 1^2 = 1 \] Finalmente, multiplicamos por 2: \[ 2 \cdot 1 = 2 \] **Resultado:** \( 2 \) --- ### c. \( \left(\frac{2}{6}\right)^{2}+\sqrt[3]{\frac{1}{27}} \) Primero, simplificamos \( \frac{2}{6} \): \[ \frac{2}{6} = \frac{1}{3} \] Ahora elevamos al cuadrado: \[ \left(\frac{1}{3}\right)^{2} = \frac{1}{9} \] Ahora resolvemos la raíz cúbica: \[ \sqrt[3]{\frac{1}{27}} = \frac{1}{3} \] Sumamos las dos partes: \[ \frac{1}{9} + \frac{1}{3} \] Convertimos \( \frac{1}{3} \) a novenos: \[ \frac{1}{3} = \frac{3}{9} \] Ahora sumamos: \[ \frac{1}{9} + \frac{3}{9} = \frac{4}{9} \] **Resultado:** \( \frac{4}{9} \) --- ### d. \( \frac{2}{5} \cdot \frac{10}{8}+\frac{1}{4} \) Primero, multiplicamos las fracciones: \[ \frac{2}{5} \cdot \frac{10}{8} = \frac{2 \cdot 10}{5 \cdot 8} = \frac{20}{40} = \frac{1}{2} \] Ahora sumamos \( \frac{1}{2} + \frac{1}{4} \). Convertimos \( \frac{1}{2} \) a cuartos: \[ \frac{1}{2} = \frac{2}{4} \] Sumamos: \[ \frac{2}{4} + \frac{1}{4} = \frac{3}{4} \] **Resultado:** \( \frac{3}{4} \) --- ### e. \( \sqrt{2 \cdot\left(\frac{22}{16}+\frac{133}{8}\right)} \) Primero, simplificamos \( \frac{22}{16} \): \[ \frac{22}{16} = \frac{11}{8} \] Ahora sumamos: \[ \frac{11}{8} + \frac{133}{8} = \frac{11 + 133}{8} = \frac{144}{8} = 18 \] Ahora multiplicamos por 2: \[ 2 \cdot 18 = 36 \] Finalmente, sacamos la raíz cuadrada: \[ \sqrt{36} = 6 \] **Resultado:** \( 6 \) --- Así que los resultados finales son

Quick Answer

a. \( \frac{16}{21} \) b. \( 2 \) c. \( \frac{4}{9} \) d. \( \frac{3}{4} \) e. \( 6 \)
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