Question

Please answer the following question

Answer 1

The IDENTITIES given in the question are used and some other identities are also used

IDENTITIES USED:

$>( (a)^{m} ) {n} = {(a)}^{mn}$

$> {a}^{ - m} = \dfrac{1}{ {a}^{m} }$

$> \dfrac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}$

Answer 2

$\bf\huge\underline\purple{Solution}$

$({ \sqrt{x} })^{ \frac{ - 2}{3} } . \sqrt{ {y}^{4} } \div \sqrt{xy \frac{ - 1}{2} }$

$= \dfrac{ ({ \sqrt{x}) }^{ \frac{ - 2}{3} } . { \sqrt{y} }^{4} }{ \sqrt{x. {y}^{ \frac{ - 1}{2} } } }$

$= \dfrac{ {x}^{ \frac{1}{2} \times \frac{ - 2}{3} }. {y}^{4 \times \frac{1}{2} } }{ {x}^{ \frac{1}{2} }. {y}^{ \frac{1}{2} \times \frac{ - 1}{2} } }$

$= \dfrac{ {x}^{ \frac{ - 1}{3} }. {y}^{2} }{ {x}^{ \frac{1}{2} }. {y}^{ \frac{ - 1}{4} } }$

$since \: \frac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}$

$= {x}^{ \frac{ - 1}{3} - \frac{1}{2} } . {y}^{2 + \frac{1}{4} }$

$= {x}^{ \frac{ - 2 - 3}{6} } . {y}^{ \frac{8 + 1}{4} }$

$= {x}^{ \frac{ - 5}{6} } . {y}^{ \frac{9}{4} }$