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

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

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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}

(x^{3} + 5)^{3} + 5

(7 x + 8)^{2}

(x^{8} - x^{4})^{3}

- 16 \cdot 9

(a - 10)^{2}

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