展开 (5^3+6)lef(t)t(4^2-3)'

解答

展开

(5^{3} + 6) l e f (t) t (4^{2} - 3)^{'}

131 e lt f (t) 13'

求解步骤

(5^{3} + 6) l e f (t) t (4^{2} - 3)^{'}

= e (5^{3} + 6) lt f (t) (4^{2} - 3)^{^{'}}

使用分配律:

a (b + c) = ab + a c

a = l e f (t) t (4^{2} - 3)^{'}, b = 5^{3}, c = 6

= l e f (t) t (4^{2} - 3)^{^{'}} 5^{3} + l e f (t) t (4^{2} - 3)^{^{'}} 6

= 5^{3} e lt f (t) (4^{2} - 3)^{^{'}} + 6 e lt f (t) (4^{2} - 3)^{^{'}}

化简

5^{3} e lt f (t) (4^{2} - 3)^{'} + 6 e lt f (t) (4^{2} - 3)^{'} : 131 e lt f (t) 1 3^{'}

= 131 e lt f (t) 1 3^{'}

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

s im pl i f y (9 - p^{2} sin^{2} (y) (cos (2 x))) p^{2} sin^{2} (y)

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

s im pl i f y (- 1 + 42) \cdot 2

e x p an d (x^{4} - 2 x^{2} + 1) - (x - 2 x = 1)