What is (2^{-1}+3^{-1})^{-1} ?

The solution to (2^{-1}+3^{-1})^{-1} is 6/5

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simplify (2^{-1}+3^{-1})^{-1}

simplify

(2^{- 1} + 3^{- 1})^{- 1}

\frac{6}{5}

Decimal Notation

1.2

Solution steps

(2^{- 1} + 3^{- 1})^{- 1}

Apply exponent rule:

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

(2^{- 1} + 3^{- 1})^{- 1} = \frac{1}{2 ^{- 1} + 3 ^{- 1}}

= \frac{1}{2 ^{- 1} + 3 ^{- 1}}

Simplify

2^{- 1} + 3^{- 1} : \frac{1}{2} + \frac{1}{3}

= \frac{1}{\frac{1}{2} + \frac{1}{3}}

Join

\frac{1}{2} + \frac{1}{3} : \frac{5}{6}

= \frac{1}{\frac{5}{6}}

Apply the fraction rule:

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

= \frac{6}{5}

1 - 1

13 a + 26 b - 39 c

- 30 \div 6

4 b - 7 - 15 b + 3

\frac{3}{5} \cdot \frac{5}{2}