What is ((2x)^{-1}+(2y)^{-1})^{-1} ?

The solution to ((2x)^{-1}+(2y)^{-1})^{-1} is (2xy)/(y+x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify ((2x)^{-1}+(2y)^{-1})^{-1}

simplify

((2 x)^{- 1} + (2 y)^{- 1})^{- 1}

\frac{2 x y}{y + x}

Solution steps

((2 x)^{- 1} + (2 y)^{- 1})^{- 1}

Apply exponent rule:

a^{- 1} = \frac{1}{a}

((2 x)^{- 1} + (2 y)^{- 1})^{- 1} = \frac{1}{( 2 x ) ^{- 1} + ( 2 y ) ^{- 1}}

= \frac{1}{( 2 x ) ^{- 1} + ( 2 y ) ^{- 1}}

Simplify

(2 x)^{- 1} + (2 y)^{- 1} : \frac{1}{2 x} + \frac{1}{2 y}

= \frac{1}{\frac{1}{2 x} + \frac{1}{2 y}}

Join

\frac{1}{2 x} + \frac{1}{2 y} : \frac{y + x}{2 x y}

= \frac{1}{\frac{y + x}{2 x y}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

\frac{24}{180}

5 \frac{3}{4} - 1 \frac{1}{8}

(\frac{3}{2})^{3}

(t - 1)^{2} - 49

\frac{50}{12}

What is ((2x)^{-1}+(2y)^{-1})^{-1} ?
The solution to ((2x)^{-1}+(2y)^{-1})^{-1} is (2xy)/(y+x)