Question

⚠️EJERCICIO 8⚠️
⚠️PASO POR PASO⚠️
NO ME VALE LA SOLUCION Y YA.
BUSCO TODOS Y PASO POR PASO !
⚠️ES EL 8, NO EL 7!⚠️

⚠️25PTS⚠️

Answer 1

Respuesta:

Pregunta 1

$\frac{3}{4}-(\frac{1}{2})^{2} \times ((\frac{5}{2})^{2}-\frac{2}{3}\times(\frac{3}{2})^{3})- \sqrt{\frac{49}{16}}$

$\frac{3}{4} - \frac{1}{4} \times (\frac{25}{4} - \frac{2}{3} \times \frac{ {3}^{3} }{ {2}^{3} }) - \frac{7}{4}$

$\frac{3}{4} - \frac{1}{4} \times ( \frac{25}{4} - \frac{9}{4} ) - \frac{7}{4}$

$\frac{3}{4} - \frac{1}{4} \times ( \frac{16}{4} ) - \frac{7}{4}$

$\frac{3}{4} - \frac{1}{4} \times 4 - \frac{7}{4} \\ \frac{3}{4} - \frac{7}{4} - 1 \\ - \frac{4}{4} - 1 \\ - 1 - 1 = - 2$

Pregunta 2

$\frac{3}{4} - ( \frac{1}{2} )^{2} \times \sqrt{ {5}^{2} - {4}^{2} } \div ( \frac{3}{2} )^{2}$

$\frac{3}{4} - \frac{1}{4} \times \sqrt{25 - 16} \div \frac{9}{4} \\ \frac{3}{4} - \frac{1}{4} \times \sqrt{9} \times \frac{4}{9} \\ \frac{3}{4} - \frac{1}{4} \times 3 \times \frac{4}{9} \\ \frac{3}{4} - \frac{1}{4} \times \frac{4}{3} \\ \frac{3}{4} - \frac{1}{3} = \frac{9 - 4}{12} = \frac{5}{12}$

Pregunta 3

${4}^{ - 1} + \sqrt{ \frac{2}{18} } \times ( {3}^{2} - {2}^{3} \times 3) \times {5}^{ - 1} \\ \frac{1}{4} + \sqrt{ \frac{1}{9} } \times (9 - 8 \times 3) \times \frac{1}{5} \\ \frac{1}{4} + \frac{1}{3} \times (9 - 24) \times \frac{1}{5} \\ \frac{1}{4} + \frac{1}{3} \times ( - 15) \times \frac{1}{5} \\ \frac{1}{4} + \frac{1 \times 1}{3 \times 5} \times ( - 15) \\ \frac{1}{4} + \frac{1}{15} \times ( - 15) \\ \frac{1}{4} + \frac{1}{1} \times - 1 \\ \frac{1}{4} - 1 = \frac{1 - 4}{4} = - \frac{3}{4}$