What is (ab^{-1}+a^{-1}b)(a^{-2}+b^{-2})^{-1} ?

The solution to (ab^{-1}+a^{-1}b)(a^{-2}+b^{-2})^{-1} is ab

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simplify (ab^{-1}+a^{-1}b)(a^{-2}+b^{-2})^{-1}

simplify

(a b^{- 1} + a^{- 1} b) (a^{- 2} + b^{- 2})^{- 1}

ab

Solution steps

(a b^{- 1} + a^{- 1} b) (a^{- 2} + b^{- 2})^{- 1}

Simplify

a b^{- 1} + a^{- 1} b : \frac{a}{b} + \frac{b}{a}

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

Simplify

(a^{- 2} + b^{- 2})^{- 1} : \frac{a ^{2} b ^{2}}{b ^{2} + a ^{2}}

= (\frac{a}{b} + \frac{b}{a}) \frac{a ^{2} b ^{2}}{b ^{2} + a ^{2}}

Apply the fraction rule:

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

= \frac{( \frac{a}{b} + \frac{b}{a} ) a ^{2} b ^{2}}{b ^{2} + a ^{2}}

Simplify

(\frac{a}{b} + \frac{b}{a}) a^{2} b^{2} : (a^{2} + b^{2}) ab

= \frac{( a ^{2} + b ^{2} ) ab}{b ^{2} + a ^{2}}

Apply the commutative law:

a^{2} + b^{2} = b^{2} + a^{2}

= \frac{( b ^{2} + a ^{2} ) ab}{b ^{2} + a ^{2}}

Cancel the common factor:

b^{2} + a^{2}

= ab

e^{- y} e^{y}

0.5 \cdot 1 0^{- 6} (10000)

(e^{x + 7})^{2}

(\frac{( 0.19 = 0.1 )}{0.8})^{- 2}

(2 x + 7)^{\circ} (6 x + 5)^{\circ}

What is (ab^{-1}+a^{-1}b)(a^{-2}+b^{-2})^{-1} ?
The solution to (ab^{-1}+a^{-1}b)(a^{-2}+b^{-2})^{-1} is ab