(1/4x^2y^3)^2+(3/4x^3y^4)×(-4xy) (a+b-c)(a+b+c)-[(a-b)^2+4ab]已知2x=y,求代数式[（x^2+y^2)-（x-y)^2+2y(x-y）]+(4y)的值

1/4x^2y^3和3/4x^3y^4这两个分数中应该只是4在分母里吧？
(1/4x^2y^3)^2+(3/4x^3y^4)×(-4xy)
=1/16x^4y^6-3x^4y^5
=x^4y^5(y/16-3)
(a+b-c)(a+b+c)-[(a-b)^2+4ab]
=[(a+b)-c][(a+b)+c]-(a^2-2ab+b^2+4ab)
=(a+b)^2-c^2-(a^2+2ab+b^2)
=(a+b)^2-c^2-(a+b)^2
=-c^2
已知2x=y，求代数式[（x^2+y^2)-（x-y)^2+2y(x-y）]+(4y)的值
[(x^2+y^2)-(x-y)^2+2y(x-y)]+(4y)
=[x^2+y^2-(x^2-2xy+y^2)+2xy-2y^2]+4y
=(x^2+y^2-x^2+2xy-y^2+2xy-2y^2)+4y
=4xy-2y^2+4y
=2y(2+2x-y)
=2y(2+y-y)
=4y

2X{SDU}14{4Y}

1)=(1/4x^2y^3)^2-3x^4y^5=1/16x^4y^5=x^4y^5(1/16y-3)
2)=(a+b)^2-c^2-(a-b)^2-4ab=a^2+2ab+b^2-c^2-a^2+2ab-b^2-4ab=-c^2
3)=(x^2+y^2-x^2+2xy-y^2+2xy-2y^2)+4y=4xy-2y^2+4y
因为y=2x
所以=2y^2-2y^2+4y=4y

(1/4x^2y^3)^2+(3/4x^3y^4)×(-4xy) (a+b-c)(a+b+c)-[(a-b)^2+4ab]不是等式啊，能算条件么？