# How do I simplify ((x^2)/(4y^-2))^(-1/2)?

##### 2 Answers

$\frac{2}{x y}$

#### Explanation:

${\left({x}^{2} / \left\{4 {y}^{- 2}\right\}\right)}^{- \frac{1}{2}}$

$= {\left(\frac{{x}^{2} {y}^{2}}{4}\right)}^{- \frac{1}{2}}$

$= {\left(\frac{{\left(x y\right)}^{2}}{4}\right)}^{- \frac{1}{2}}$

$= \frac{{\left({\left(x y\right)}^{2}\right)}^{- \frac{1}{2}}}{{4}^{- \frac{1}{2}}}$

$= {4}^{\frac{1}{2}} {\left(x y\right)}^{2 \left(- \frac{1}{2}\right)}$

$= {\left({2}^{2}\right)}^{\frac{1}{2}} {\left(x y\right)}^{- 1}$

$= \setminus \frac{{2}^{2 \left(\frac{1}{2}\right)}}{x y}$

$= \frac{2}{x y}$

Jul 26, 2018

${\left({x}^{2} / \left(4 {y}^{- 2}\right)\right)}^{- \frac{1}{2}} = \frac{2}{x y}$

#### Explanation:

Let ,

$A = {\left({x}^{2} / \left(4 {y}^{- 2}\right)\right)}^{- \frac{1}{2}}$

$\therefore A = {\left({x}^{2}\right)}^{- \frac{1}{2}} / \left({\left(4 {y}^{- 2}\right)}^{- \frac{1}{2}}\right) \ldots . . \to \left[\because {\left(\frac{a}{b}\right)}^{n} = {a}^{n} / {b}^{n}\right]$

$\therefore A$=${x}^{2 \times \left(- \frac{1}{2}\right)} / \left({4}^{- \frac{1}{2}} \cdot {y}^{\left(- 2\right) \left(- \frac{1}{2}\right)}\right) \to \left[\because {\left({a}^{m}\right)}^{n} = {a}^{m n} , {\left(a b\right)}^{n} = {a}^{n} \cdot {b}^{n}\right]$

:.A=x^(-1)/((2^2)^(-1/2)y^1

$\therefore A = \frac{\frac{1}{x}}{{2}^{-} 1 y}$

$\therefore A = {2}^{1} \times \frac{1}{x y}$

$\therefore A = \frac{2}{x y}$