x y=36 25乘以2x=40y

把y=3/x代入(x+y)乘以根号下x/y原式=(x+3/x)乘以根号下x^2/3=(x^2+3)/x乘以根号3x的绝对值/3当x》0时原式=根号3(x^2+3)/3当x

把13X^2-6XY+Y^2-4x+1=0配方,得13X^2-6XY+Y^2-4x+1=09x^2-6xy+y^2+4x^2-4x+1=0(3x-y)^2+(2x-1)^2=0又因为(2x-1)^2>

由x-3y=0可知x=3y原式=(2x+y)(x-y)/(x-y)²=(2x+y)/(x-y)=(6y+y)/(3y-y)=7y/2y=7/2

x/y=3所以x=3y(x²+2xy-3y²)/(x²-xy+y²)=(9y²+6y²-3y²)/(9y²-3y&sup

(x-y)²=(x+y)²-4xy=36+8=44x-y=±√44=±2√11于是（x^4-y^4）除以x^2+y^2/xy乘以（6x+6y）=（x²+y²)(

2X-Y=0,2x=y(2X+Y)/(X^2-2XY+Y^2)×（X-Y）=4x/(x^2-4x^2+4x^2)×(-x)=-4

x^3y-2x^2y^2+xy^3=x^3y+2x^2y^2+xy^3-4x^2y^2=xy(x^2+2xy+y^2)-4x^2y^2=xy(x+y)^2-4x^2y^2=12×7×7-4×12×12

式子是这样么(2x+y)/(x^2-2xy+y^2）=(2x+y)/(x-y)^2x-3y=0x=3y代入原式(2*3y+y)/(3y-y)^2=7/4yy不能等于0

XY(XY-X^2)(X-Y)=X^2Y(X-Y)(X-Y)=(X-Y)(X^2Y-1)再问：答案是。-X2Y.我不知道过程是如何来的。能告诉我过程吗。谢谢。负X的平方Y

对.前提是x不等于y

a-b=-xy+3y^2-x^2-3a+2b=42x^2-4y^2

x-3y=0x=3y原式=(2x+y)(x-y)/(x-y)²=(2x+y)/(x-y)=(6y+y)/(3y-y)=7y/2y=7/2

(x-xy)+(xy-y)=40-20x-xy+xy-y=20x-y=20(x-xy)-(xy-y)=40-(-20)x-xy-xy+y=60x+y-2xy=60

x^2+y^2+5/4=2x+y(x^2-2x+1)+(y^2-y+1/4)=0(x-1)^2+(y-1/2)^2=0平方大于等于0,相加等于0,若有一个大于0,则另一个小于0,不成立所以两个都等于0

A=8x的二次方乘以y-6xy的二次方-3xyB=7xy的二次方-2xy+5x的二次方乘以yA+B=13x的二次方乘以y+xy的二次方-5xy=3CC=(13x的二次方乘以y+xy的二次方-5xy)/

2+x^2+(x^2)*(y^2)=2X-2xy,化简得：（x-1)^2+(xy-1)^2=0【把化简前式子中的2分成1+1】所以x-1=0,xy-1=0,得x=y=1【（x-1)^2是（x-1)的平

x^2+4y^2-4x+4y+5=0,(x-2)^2+(2y+1)^2=0,x=2,y=-1/2,(2x^2+xy-y^2分之x^4-y^4)*（xy-y^2分之2x-y)/[(x^2+y^2)/y]

x^2+4y^2-2x+4y+5=0==>(x-1)^2+(2y+1)^3+3=0无解?还是要用到虚数什么的

x^2+4y^2-4x+4y=5=0?如果条件是x^2+4y^2-4x+4y+5=0,那么变形就有x^2-4x+4+4y^2+4y+1=0,即（x-2）^2+(2y+1)^2=0,很显然x=2,y=-

x-y=(x-xy)+(xy-y)=40+(-20)=20x+y-2xy=(x-xy)-(xy-y)=60