展开 λe^{λt}g(e^t-λe^t)

解答

展开

λ e^{λ t} g (e^{t} - λ e^{t})

λ e^{t + λ t} g - λ^{2} e^{t + λ t} g

求解步骤

λ e^{λ t} g (e^{t} - λ e^{t})

使用分配律:

a (b - c) = ab - a c

a = λ e^{λ t} g, b = e^{t}, c = λ e^{t}

= λ e^{λ t} g e^{t} - λ e^{λ t} g λ e^{t}

= λ e^{t} e^{λ t} g - λλ e^{t} e^{λ t} g

化简

λ e^{t} e^{λ t} g - λλ e^{t} e^{λ t} g : λ e^{t + λ t} g - λ^{2} e^{t + λ t} g

= λ e^{t + λ t} g - λ^{2} e^{t + λ t} g

e x p an d \frac{15 a}{s ( 3 s + 8 )}

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

e x p an d x (1 + x ln (x y - 5))

e x p an d \frac{5}{6} - 2

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