已知,用4x除(ax^3+bx^2-4cx)的商式为x^2+2x-c,且c不等于0,求(3ac^2-bc)除以c-3c(ac-ab^2)除以c的值在现等的挖.知道到的告诉偶哈...thanks`

正确的答案如下：

(ax^3+bx^2-4cx)=4x*(x^2+2x-c)
(ax^3+bx^2-4cx)=4x^3+8x^2-4cx
左右对照，得
a=4
b=8
所以
(3ac^2-bc)÷c-3c(ac-ab^2)÷c
=(3ac-b)-3(ac-ab^2)
=3ac-b-3ac+3ab^2
=b(3ab-1)
=8*(3*4*8-1)
=8*95
=760

(ax^3+bx^3-4cx)/4x=(ax^2+bx^2-4c)/4=x^2+2x-c 所以a=4,b=8 所以(3ac^2-bc)除以c-3c(ac-ab^2)除以c=[(3ac^2-bc)/c]-[3c(ac-ab^2)]/c =[3ac^2-bc-3ac^2+3ab^2c]/c =[3ab^2c-bc]/c =3ab^2-b =3*4*8^2-8 =760