When simplified, $(x^{- 1} + y^{- 1})^{- 1}$ equals

x+y

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\frac{x y}{x + y}

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\frac{x + y}{x y}

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\frac{1}{x + y}

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Solution

The correct option is B $\frac{x y}{x + y}$
$(x^{- 1} + y^{- 1})^{- 1} = {(\frac{1}{x} + \frac{1}{y})}^{- 1} = {(\frac{y + x}{x y})}^{- 1} = \frac{x y}{x + y}$

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Similar questions

[\frac{x + y}{x^{- 1} + y^{- 1}}] - x y

When simplified $(x^{- 1} + y^{- 1})^{- 1}$ is equal to

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= π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

{t a n}^{- 1} \sqrt{{\frac{x (x + y + z)}{y z}}} + {t a n}^{- 1} \sqrt{{\frac{y (x + y + z)}{z x}}} + {t a n}^{- 1} \sqrt{{\frac{z (x + y + z)}{x y}}} =

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\neq

\neq

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