展开 e^{3x}*(u_{2}(t)(t^3+1))

解答

展开

e^{3 x} \cdot (u_{2} (t) (t^{3} + 1))

e^{3 x} u_{2} t^{4} + e^{3 x} u_{2} t

求解步骤

e^{3 x} (u_{2} (t) (t^{3} + 1))

去除括号:

(a) = a

= e^{3 x} u_{2} t (t^{3} + 1)

使用分配律:

a (b + c) = ab + a c

a = e^{3 x} u_{2} t, b = t^{3}, c = 1

= e^{3 x} u_{2} t t^{3} + e^{3 x} u_{2} t \cdot 1

= e^{3 x} u_{2} t^{3} t + 1 \cdot e^{3 x} u_{2} t

化简

e^{3 x} u_{2} t^{3} t + 1 \cdot e^{3 x} u_{2} t : e^{3 x} u_{2} t^{4} + e^{3 x} u_{2} t

= e^{3 x} u_{2} t^{4} + e^{3 x} u_{2} t

e x p an d (3 x^{2} - 2) e^{x^{3} + 7 x}

e x p an d \frac{11 a ^{3} - 9 a}{( 3 a - 2 ) ( 3 + 2 a )}

\frac{\partial}{\partial t} ((u (t, x)) (3 x - 2 t))

e x p an d - \frac{d}{d s} (\frac{a}{s ( s ^{2} - b )})

e x p an d - \frac{12 ( 6 + 2 )}{17}