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

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

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

simplify

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

\frac{1}{e ^{2} + e + 1}

Decimal Notation

0.09003 \dots

Solution steps

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

Apply exponent rule:

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

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

e^{1 - 1} + e^{2 - 1} + e^{3 - 1} = 1 + e + e^{2}

= \frac{1}{1 + e + e ^{2}}

A t e^{- s t}

18 x^{2} y^{2} z^{4}

e^{- (x - k)^{2}}

x^{4} (x^{4} sin^{4} (y))

2 e \cdot e^{3}

What is (e^{1-1}+e^{2-1}+e^{3-1})^{-1} ?
The solution to (e^{1-1}+e^{2-1}+e^{3-1})^{-1} is 1/(e^2+e+1)