若x²+2y²-2xy+4y﹢4=0

对.前提是x不等于y

(-3x^y+2xy)-(4x^+xy)=-3x^y+2xy-4x^-xy=-3x^y+xy-4x^所以填上-3x^y+xy-4x^

4xy-(x+y)^2/xy(x+y)(x-y)÷x^2+xy-2y^2/x^2y+2xy^2=(4xy-x²-2xy-y²)/[xy(x+y)(x-y)]÷[(x-y)(x+2y

是不是-2A+B=2a所以-4x²+6xy-2y²+2x-4y+4x²-6xy+2y²+3x-y=5x-5y=5a+15=2aa=-3A=18-27+9+3-6

xy-[x+2y+(0.5x-2)×(y-2)]=24xy-[x+2y+0.5xy-x-2y+4]=24xy-x-2y-0.5xy+x+2y-4=24xy-0.5xy=24+40.5xy=28xy=5

【x²+xy/（x-y）】/【xy/（x-y）】=【x²（x-y）/（x-y）+xy/（x-y）】/【xy/（x-y）】={【x²（x-y）+xy】/（x-y）}/【xy

(x*x+y*y)(x*x+y*y)-4x*x*y*y=(x^4-2x^2y^2+y^4)=(x^2-y^2)^2=0x^2=y^2x/y=±1(x*x+5xy+y*y)/(x*x+2xy+y*y)=

x^3y(-4y)^2+(-7xy)^2*(-xy)-5xy^3*(-3x)^2=16x^3y^3-49x^3y^3-45^3y^3=-78x^3y^3如果本题有什么不明白可以追问,

(x^-4y^)/(x^+2xy+y^)/(x+2y)/(2x^+2xy)=1.=3(a-b)/(10ab)*(25a^2b^3)/((a+b)(a-b))=15/2*(ab^2)/(a+

原式=-(x+2y)*(x^2-2xy+4y^2)=-(x^3+(2y)^3)=-x^3-8y^3

(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-2xy+2x+3y-3xy-2y+2x-x-4y-xy=-6xy+3x-3y=-6xy+3*(x-y)当时,原式=-6*1+3*

即(x-2y)²=0x-2y=0所以x=2y所以原式=(2x²+2xy-xy-y²)/(4x²-4xy+y²)=(2x²+xy-y²

设A=2x²-3xy+y²-x+2y,B=4x²-6xy+2y²-3x-y,且B-2A=a所以a=B-2A=4x²-6xy+2y²-3x-y

=(x^4+x^3y)+(y^4+xy^3)+(4x^3y+4x^2y^2)+(4xy^3+4x^2y^2)+x^2y+xy^2=x^3(x+y)+y^3(x+y)+4x^2y(y+x)+4xy^2(

∵x-y=4xy，∴2x+3xy-2yx-2xy-y=2(x-y)+3xyx-y-2xy=8xy+3xy4xy-2xy=112．故答案为：112．

∣x-2a∣+（y+3）²=0因为∣x-2a∣≥0,（y+3）²≥0所以∣x-2a∣=0,（y+3）²=0所以x=2a,y=-3B-2A=(4x²-6xy+2y

(2x-14xy-2y)除以(x-2xy-y)=4(2x-14xy-2y)=4(x-2xy-y)2x+6xy-2y=0x-y=-3xy两边同除以xy,得1/y-1/x=-3所以1/x-1/y=3

x^2y+4xy+4y=y(x²+4x+4)=y(x+2)²

得y-x=3xyx-y=-3xy原式=-15xy+2xy/-3xy-4xy=-13/-7=13/7

[-(x-y)^2/xy]^-4*(y^2-xy/x)^3*(x^4/y^10)÷(xy-y^2/x)^-5=-y(y-x)^29^-x*27^y=3^-2x*3^3y=3^(-2x+3y)=3^(-