展开 (b+bc)(b+b^'c)(b+d)

解答

展开

(b + b c) (b + b^{^{'}} c) (b + d)

b^{3} + b^{2} d + c b^{' + 2} + c d b^{1 +'} + b^{3} c + b^{2} c d + c^{2} b^{' + 2} + c^{2} d b^{1 +'}

求解步骤

(b + b c) (b + b^{^{'}} c) (b + d)

= (b + b c) (b + c b^{^{^{'}}}) (b + d)

乘开

(b + b c) (b + b^{^{'}} c) : b^{2} + c b^{1 +^{'}} + b^{2} c + c^{2} b^{1 +^{'}}

= (b^{2} + c b^{1 +^{'}} + b^{2} c + c^{2} b^{1 +^{'}}) (b + d)

乘开

(b^{2} + c b^{1 +^{'}} + b^{2} c + c^{2} b^{1 +^{'}}) (b + d) : b^{3} + b^{2} d + c b^{^{'} + 2} + c d b^{1 +^{'}} + b^{3} c + b^{2} c d + c^{2} b^{^{'} + 2} + c^{2} d b^{1 +^{'}}

= b^{3} + b^{2} d + c b^{^{'} + 2} + c d b^{1 +^{'}} + b^{3} c + b^{2} c d + c^{2} b^{^{'} + 2} + c^{2} d b^{1 +^{'}}

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

e x p an d A (- 2, - 1) B (- 4, 1) c (- 1, 4)

s im pl i f y (x + \frac{1}{4}) (x - 5)

e x p an d (x^{3} + 4 x^{2}) + (3 x^{2} + 12 x) i

e x p an d \frac{( 1 - e ^{- a t} )}{( z - 1 ) ( z - e ^{- a t} )}