x除以y=712乘2xy成什么比例

由X+Y=3XY可得X+Y-XY=2XY,用X+Y-XY替换分母中的2XY,所以(2X+2Y-XY)/(X+Y+2XY)=(2X+2Y-XY)/(2X+2Y-XY)=1

原式=[(x+2y)/(x+y)][xy/(x+2y)]÷[(x+y)/xy]=[xy/(x+y)]×xy/(x+y)=x²y²/(x+y)²

[8x^3y^3-(-xy^2)^2]/a=-x^2y^2+1/8*xy^3a=(8x^3y^3-x^2y^4)/(1/8*xy^3-x^2y^2)=-8xya/4x=-2y

1.已知x.y满足x^2+y^2-4x-6y+13=0,（x-2)^2+(y-3)^2=0x=2,y=3(-y/x^3)^3除以(-1/xy)^4乘(x/y^2)^2=(-3/2^3)^3÷（-1/2

(xy-x的平方）除以xy分之x的平方-2xy+y的平方乘x的平方分之x-y=x(y-x)*xy/(x-y)²*（x-y）/x²=-y再问：不好意思，打错了(xy-y的平方）除以x

(xy-x²)÷(x²-2xy+y²)/(x-y)×(x-y)/x²=-x(x-y)÷(x-y)²/x²=-x(x-y)×x²/(

由于xy除以x+y=2故xy=2（x+y）将这代入8x-5xy+8y除以-x+3xy-y化简得其中的（x+y）会相互约去故8x-5xy+8y除以-x+3xy-y=-2/5希望这个能帮到你over~

∵x^2+y^2-2x+4y+5=0配方：∴(x-1)^2+(y+2)^2=0那么x-1=0且y+2=0所以x=1,y=-2∴(x^4-y^4)/(2x^2+xy-y^2)*（2x-y)/(xy-y^

(1)（xy-x²)除以xy分之x²+y²-2xy乘x²分之x-y=x(y-x)×xy/(x-y)²×(x-y)/x²=-y(2)先化简：（

原式=[4(x+y)-2xy]分之[(x+y)+xy]=[4(3xy)-2xy]分之[(3xy)+xy]=10xy分之2xy=5分之1

原式=2x(xy²-x²)÷(-2xy)=(x²-xy²)/y望楼主采纳哦~~~

(3a^2/x+2y)^3乘(x^2-4xy+4y^2)除以(2y-x/x+2y)^2=27a^6/(x+2y)^3*(x-2y)^2÷(2y-x)^2/(x+2y)^2=27a^6/(x+2y)

带入x=2,y=1,最后得出-4

X²+2xy+y²/xy乘x²-2xy+y²/xy+y²=(x+y)²/xy×(x-y)²/y(x+y)=(x+y)(x-y)&#

4.5÷x=4÷y4.5y=4xy=8/9*x所以x和y成正比例

8x=7/y可转化为x=7/8y所以,随着y的增大,x逐渐减小,所以x、y成反比例关系再问：好那我做对了，这次考试满分，有一定的可能性了。再答：哦，原来是个数学尖子呀，幸亏我学习不错，要不就要被你鄙视

xy^2*(-x^2y^2)(1/2xy^3)^2=-xy^2*x^2y^2*1/4x^2y^6=-1/4x^5y^10(xy^2)^5=x^5y^10当xy^2=2时,原式=-1/4*2^5=-1/

是这样的形式吗?[x/（x+y）+2y/（x+y）]×[xy/（x+y）]÷（1/x+1/y）一:原式=（x+2y）/（x+y）×[xy/（x+2y）]÷[（x+y）/xy]=x²y&sup

（-y/x^2)^3除以（-1/xy)^4乘（x/y^2)^2=-y^3/x^6÷(1/x^4y^4)×x^2/y^4=-y^3/x^6×x^4y^4×x^2/y^4=-y^3x^2+y^2-4x-6

因为x²+4y²+x²y²-6xy+1=0(x²-4xy+4y²)+(x²y²-2xy+1)=0(x-2y)²