Question

APLICACIÓN de La ley de "sandwich" extremos y medios. ​

Answer 1

Explicación paso a paso:

b.

$\frac{ \frac{5}{6} \div ( - \frac{1}{2} + 1 + \frac{2}{3}) }{ \frac{4}{3} + \frac{1}{6} } \\ \\ = \frac{ \frac{5}{6} \div \frac{ - 3 + 6 + 4}{6} }{ \frac{8 + 1}{6} } \\ \\ = \frac{ \frac{5}{6} \div \frac{7}{6} }{ \frac{9}{6} } \\ \\ = \frac{ \frac{5}{6} \times \frac{6}{7} }{ \frac{9}{6} } \\ \\ = \frac{ \frac{5}{7} }{ \frac{9}{6} } \\ \\ = \frac{5 \times 6}{7 \times 9} \\ \\ = \frac{10}{21}$

c.

$\frac{ \frac{9}{13} + \frac{1}{ \frac{1}{4} + 3 } + 2 }{ - \frac{2}{ - \frac{1}{2} + \frac{5}{3} } + 3 }$

resolvemos el numerador:

$\frac{9}{13} + \frac{1}{ \frac{1}{4} + 3 } + 2 \\ \\ = \frac{9}{13} + \frac{1}{ \frac{13}{4} } + 2 \\ \\ = \frac{9}{13} + \frac{1 \times 4}{ 13} + 2 \\ \\ = \frac{9}{13} + \frac{4}{ 13} + 2 \\ \\ =\frac{9 + 4}{13} + 2 \\ \\ = \frac{13}{13} + 2 \\ \\ = 1 + 2 \\ = 3$

ahora resolvemos el denominador:

$- \frac{2}{ - \frac{1}{2} + \frac{5}{3} } + 3 \\ \\ = - \frac{2}{ \frac{ - 3 + 10}{6} } + 3 \\ \\ = - \frac{2}{ \frac{ 7}{6} } + 3 \\ \\ = - \frac{2 \times 6}{7} + 3 \\ \\ = \frac{ - 12}{7} + 3 \\ \\ = \frac{9}{7}$

Finalmente tenemos:

$= \frac{3}{ \frac{9}{7} } \\ \\ = \frac{3 \times 7}{9} \\ \\ = \frac{7}{3} = 2 \frac{1}{3}$