展开 ((1-λ)q+λp)(a*e^{b(1-λ)q+λp}+c)

解答

展开

((1 - λ) q + λ p) (a \cdot e^{b (1 - λ) q + λ p} + c)

a e^{b q - λb q + λ p} q - λa e^{b q - λb q + λ p} q + c q - λ c q + λ p a e^{b q - λb q + λ p} + λ p c

求解步骤

((1 - λ) q + λ p) (a e^{b (1 - λ) q + λ p} + c)

= (q (1 - λ) + λ p) (a e^{b q (1 - λ) + λ p} + c)

使用 FOIL 方法:

(a + b) (c + d) = a c + a d + b c + b d

a = (1 - λ) q, b = λ p, c = a e^{b (1 - λ) q + λ p}, d = c

= (1 - λ) q a e^{b (1 - λ) q + λ p} + (1 - λ) q c + λ p a e^{b (1 - λ) q + λ p} + λ p c

= a e^{b q (1 - λ) + λ p} q (1 - λ) + c q (1 - λ) + λ p a e^{b q (1 - λ) + λ p} + λ p c

化简

a e^{b q (1 - λ) + λ p} q (1 - λ) + c q (1 - λ) + λ p a e^{b q (1 - λ) + λ p} + λ p c : a e^{b q - λb q + λ p} q - λa e^{b q - λb q + λ p} q + c q - λ c q + λ p a e^{b q - λb q + λ p} + λ p c

= a e^{b q - λb q + λ p} q - λa e^{b q - λb q + λ p} q + c q - λ c q + λ p a e^{b q - λb q + λ p} + λ p c

e x p an d (x - e) (x - 23)

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

e x p an d \frac{1}{( D ^{2} + a ^{2} ) \cdot sin ( a ) x}

e x p an d \frac{1}{x ^{\frac{4}{3}} \cdot ( e ^{2 x} + 1 )}

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