展开 e^{(t-1)}U(t-1)

解答

展开

e^{(t - 1)} U (t - 1)

e^{t - 1} t U - e^{t - 1} U

求解步骤

e^{(t - 1)} U (t - 1)

使用分配律:

a (b - c) = ab - a c

a = e^{(t - 1)} U, b = t, c = 1

= e^{(t - 1)} U t - e^{(t - 1)} U \cdot 1

= e^{(t - 1)} t U - 1 \cdot e^{(t - 1)} U

乘以:

1 \cdot e^{t - 1} = e^{t - 1}

= e^{t - 1} t U - e^{t - 1} U

e x p an d - \frac{1}{6} (x - 3 \cdot (x^{2} e^{- x}))

e x p an d \frac{n + 5}{( n ^{2} + 6 n + 8 ) ( n ^{2} + 2 n )}

e x p an d \frac{1}{( x + 1 ) ( x ^{2} - 8 s ^{17} )}

e x p an d \frac{4 x ^{4} + 2 x + 1}{( x - 1 ) ( x ^{2} + 1 )}

e x p an d x^{2} (cos (2 x + 3 y) (2 x + 3)) + sin (2 x + 3 y)