Assuming that $x, y, z$ are positive real numbers, simplify each of the following:$ (\sqrt{x})^{-2 / 3} \sqrt{y^{4}} \p \sqrt{x y^{-1 / 2}} $

Given:

\( (\sqrt{x})^{-2 / 3} \sqrt{y^{4}} \div \sqrt{x y^{-1 / 2}} \)

To do:

We have to simplify the given expression.

Solution:

We know that,

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

$a^{m} \times a^{n}=a^{m+n}$

$a^{m} \div a^{n}=a^{m-n}$

$a^{0}=1$

Therefore,

$(\sqrt{x})^{-2 / 3} \sqrt{y^{4}} \div \sqrt{x y^{-1 / 2}}=(x^{\frac{1}{2}})^{\frac{-2}{3}} \times (y^{\frac{1}{2}})^{4} \div x^{\frac{1}{2}}(y^{\frac{-1}{2}})^{\frac{1}{2}}$

$=x^{\frac{1}{2}\times(\frac{-2}{3})} \times y^{\frac{1}{2} \times 4} \div x^{\frac{1}{2}} \times y^{\frac{-1}{2} \times \frac{1}{2}}$

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

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

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

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

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

Hence, $(\sqrt{x})^{-2 / 3} \sqrt{y^{4}} \div \sqrt{x y^{-1 / 2}}=\frac{y^{\frac{9}{4}}}{x^{\frac{5}{6}}}$. 

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

44 Views

Kickstart Your Career

Get certified by completing the course

Get Started