展开 (t+2)2^{et}

解答

展开

(t + 2) 2^{e t}

2^{e t} t + 2^{1 + e t}

求解步骤

(t + 2) \cdot 2^{e t}

= 2^{e t} (t + 2)

使用分配律:

a (b + c) = ab + a c

a = 2^{e t}, b = t, c = 2

= 2^{e t} t + 2^{e t} \cdot 2

= 2^{e t} t + 2 \cdot 2^{e t}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2 \cdot 2^{e t} = 2^{1 + e t}

= 2^{e t} t + 2^{1 + e t}

e x p an d \frac{1}{( 1 - X ) ( B - X ) ( D - X )}

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

e x p an d 4 c - 3

e x p an d 2 e^{2} \cdot (e^{x} + 2 x)

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