What is (a+b)^{-1}*(a^{-1}+b^{-1}) ?

The solution to (a+b)^{-1}*(a^{-1}+b^{-1}) is 1/(ab)

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simplify (a+b)^{-1}*(a^{-1}+b^{-1})

simplify

(a + b)^{- 1} \cdot (a^{- 1} + b^{- 1})

\frac{1}{ab}

Solution steps

(a + b)^{- 1} (a^{- 1} + b^{- 1})

Apply exponent rule:

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

(a + b)^{- 1} = \frac{1}{a + b}

= \frac{1}{a + b} (a^{- 1} + b^{- 1})

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

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

= \frac{1}{a + b} (\frac{1}{a} + \frac{1}{b})

Apply the fraction rule:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot ( \frac{1}{a} + \frac{1}{b} )}{a + b}

Simplify

1 \cdot (\frac{1}{a} + \frac{1}{b}) : \frac{b + a}{ab}

= \frac{\frac{b + a}{ab}}{a + b}

Apply the fraction rule:

\frac{\frac{a}{b}}{c} = \frac{a}{b \cdot c}

= \frac{b + a}{ab ( a + b )}

Apply the commutative law:

b + a = a + b

= \frac{a + b}{ab ( a + b )}

Cancel the common factor:

a + b

= \frac{1}{ab}

\frac{1}{sin ^{2} ( a )} - \frac{1}{tan ^{2} ( a )}

(7 v - 9) - 2 (6 v + 9)

\frac{a ^{20}}{a ^{15}}

a - 1 x^{2} - 2 x + 1

(2 x^{5} y^{- 5})^{- 3}

What is (a+b)^{-1}*(a^{-1}+b^{-1}) ?
The solution to (a+b)^{-1}*(a^{-1}+b^{-1}) is 1/(ab)