初一的数学1.若 2x+y=0 , 则代数式4x^3+2xy[x+y]+y^3=[ ]2.1/4x^3y^2*[-2xy

1.若 2x+y=0 ,则代数式4x^3+2xy[x+y]+y^3
=4x^3+2x^2y+2xy^2+y^3
=x^2(2x+y)+2y^2(2x+y)
=0+0
=0
2.1/4x^3y^2*[-2xy^2]+[-2x^2y]*[-1/2xy]*3x^2y^2z
=-1/2x^4y^4+x^5y^4z
3.[3x^2+1/2y-2/3y^2]*[-1/2xy]^3
=[3x^2+1/2y-2/3y^2]*[-1/8x^3y^3]
=3x^2*[-1/8x^3y^3]+1/2y*[-1/8x^3y^3]-2/3y^2*[-1/8x^3y^3]
=-3/8x^5y^3-1/16x^3y^4+1/12x^3y^5
4.12ab{2a-3/4[a-b]+2/3b}
=12ab(2a-3/4a+3/4b+2/3b)
=12ab(5/4a+17/12b)
=12ab*5/4a+12ab*17/12b
=15a^2b+17ab^2
5.[-a]^3*[-2ab^2]^3-4ab^2*[7a^5b^4-1/2ab^3-5]
=-a^3*(-8a^3b^6)-28a^6b^6+2a^2b^5+20ab^2
=8a^6b^6-28a^6b^6+2a^2b^5+20ab^2
=-20a^6b^6+2a^2b^5+20ab^2
6.已知ab^2=6,
ab[a^2b^5-ab^3-b]
=ab[(ab^2)^2*b-(ab^)*b-b]
=ab[36b-6b-b]
=ab*29b
=29ab^2
=29*6
=174
7.若 x=1/2,y=1
x[x^2+xy+y^2]-y[x^2+xy+y^2]+3xy[y-x]
=x^3+x^2y+xy^2-x^2y-xy^2-y^3+3xy^2-3x^2y
=x^3-y^3+3xy^2-3x^2y
=1/8-1+3/2-3/4
=-1/8

你写的有点乱