4y*4y-(x*x y) (x*x-4y*4y)其中x=-28,y=18-搜答案-搜搜考试网

(-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

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

原式=(2x-y)(x+y)/(2x-y)^2=(x+y)/(2x-y)x/y=2x=2y原式=3y/3y=1

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+

(-x-2y)(x²-2xy+4y²)

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

3x^2+xy-2y^2=0推出(3x-2y)(x+y)=0推出x=-y或x=(2/3)y{(x+y)/(x-y)+4xy/(y^2-x^2)}/{(x+3y)*(x-y)}/x^2-9y^2推出：{

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

把第一个式子分开来看,x的平方-8x+16+y的平方+6Y+9=0,然后就有两个完全平方了,（x-4）的平方+（y+3）的平方=0然后可以得出x=4,y=-3,然后带进去就可以得出答案

xy+1/xy+y/x+x/y=[(xy)^2+1+x^2+y^2]/(xy)=[(xy)^2-2xy+1+x^2-2xy+y^2+4xy]/(xy)=[(xy-1)^2+(x-y)^2+4xy]/(

=[-(X-Y)/XY(X=Y)]*[XY/(X-Y)]=-1/(X+Y)=-1/25

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

x^5+y^5-(x^4y+xy^4)

x^5+y^5-(x^4y+xy^4)=(x^5－x⁴y)－(xy⁴－y^5)=x⁴(x－y)－y⁴(x－y)=(x⁴－y⁴)(

x^2y+4xy+4y

x^2y+4xy+4y=y(x²+4x+4)=y(x+2)²

再答：

题目好像有问题再问：没有问题的

[-(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^(-

原式=[(x-y)^2/(x-y)+4xy/(x-y)][(x+y)^2/(x+y)-4xy/(x+y)]={[(x-y)^2+4xy]/(x-y)}{[(x+y)^2-4xy]/(x+y)}=[(x

原式={[(x-y)^2+4xy]/(x-y)}{[(x+y)^2-4xy]/x+y}={[x^2-2xy+y^2+4xy]/(x-y)}{[x^2+2xy+y^2-4xy]/x+y}={[x^2+2