展开 (4+t)^3(e^t+e^tt)

解答

展开

(4 + t)^{3} (e^{t} + e^{t} t)

e^{t} t^{4} + 13 e^{t} t^{3} + 60 e^{t} t^{2} + 112 e^{t} t + 64 e^{t}

求解步骤

(4 + t)^{3} (e^{t} + e^{t} t)

使用分配律:

a (b + c) = ab + a c

a = (4 + t)^{3}, b = e^{t}, c = e^{t} t

= (4 + t)^{3} e^{t} + (4 + t)^{3} e^{t} t

= e^{t} (4 + t)^{3} + e^{t} t (4 + t)^{3}

化简

e^{t} (4 + t)^{3} + e^{t} t (4 + t)^{3} : e^{t} t^{4} + 13 e^{t} t^{3} + 60 e^{t} t^{2} + 112 e^{t} t + 64 e^{t}

= e^{t} t^{4} + 13 e^{t} t^{3} + 60 e^{t} t^{2} + 112 e^{t} t + 64 e^{t}

e x p an d \frac{130}{( x ^{2} + 1 ) ( x ^{2} - 25 )}

e x p an d (f (x) = 0.5 x^{2} - k (x) x - 8 k (x) (x)^{3} x - k (x)) k (x)

e x p an d \frac{x + 3}{x ( x + 8 )}

e x p an d (2 sin (x) - 1) (sin (x) + 1)

s im pl i f y - 2 (- 2^{\frac{N}{2}} + 1)