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

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

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

simplify

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

\frac{2 x}{x + 2}

Solution steps

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

Apply exponent rule:

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

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

= \frac{1}{2 ^{- 1} + x ^{- 1}}

Simplify

2^{- 1} + x^{- 1} : \frac{1}{2} + \frac{1}{x}

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

Join

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

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

Apply the fraction rule:

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

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

(\frac{1}{4})^{- \frac{5}{2}}

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