What is expand (1-e^{-λ(t+x)})/(1-e^{-λt)} ?

The solution to expand (1-e^{-λ(t+x)})/(1-e^{-λt)} is (1-e^{-λt-λx})/(1-e^{-λt)}

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expand (1-e^{-λ(t+x)})/(1-e^{-λt)}

expand

\frac{1 - e ^{- λ (t + x)}}{1 - e ^{- λ t}}

\frac{1 - e ^{- λ t - λ x}}{1 - e ^{- λ t}}

Solution steps

\frac{1 - e ^{- λ (t + x)}}{1 - e ^{- λ t}}

Expand

- λ (t + x) : - λ t - λ x

= \frac{1 - e ^{- λ t - λ x}}{1 - e ^{- λ t}}

(x - 2) \cdot e^{x}

\frac{1}{( s + 2 ) ( s + 4 ) ( s + 6 )}

9 (6 + 26)

15 (2 + lo g_{10} (48))

(t + z cos (t)) d t + (sin (t) - 3 z^{2} + 5) d z

What is expand (1-e^{-λ(t+x)})/(1-e^{-λt)} ?
The solution to expand (1-e^{-λ(t+x)})/(1-e^{-λt)} is (1-e^{-λt-λx})/(1-e^{-λt)}