若x y=10 ,xy= 1,则x三次方y

1.2(Xy+Xy)-3(Xy-xy)-4Xy=2*2xy-0-4xy=4xy-4xy=02.1/2ab-5aC-(3acb)+(3aC-4aC)=1/2ab-5ac-3acb-ac=1/2ab-6a

[（xy+2）（xy-2）-2x2y2+4]÷（xy）=（x2y2-4-2x2y2+4）÷xy=-x2y2÷xy=-xy,当x=10,y=1/25时,原式=-2/5．

[(xy+2)(xy-2)-2x^2y^2+4]/(xy)=[(x^2y^2-4)-2x^2y^2+4]/(xy)=-x^2y^2/xy=-xy=-10*(-1/25)=2/5

由x平方+xy-2x+y=0（1）,y平方+xy+x-2y=0,（2）,用（1）+（2）得：x平方+2xy+y平方-x-y=0,（x平方+2xy+y平方）-（x+y）=0（x+y)^2-(x+y)=0

1/x+1/y=2y+x=2xy(2x-xy+2y)=2(x+y)-xy=3xy(3x+5xy+3y)=3(x+y)+5xy=11xy(2x-xy+2y)÷(3x+5xy+3y)=3xy/11xy=3

x²+y²=(x+y)²-2xy=10²+2=102

由题设可知y=xy-1，∴x=yx3y=x4y-1，∴4y-1=1，故y=12，∴12x=x，解得x=4，于是x+y=4+12=92．故答案为：92．

你确定没写错题目?后面的式子一约就变成前面的式子了,答案也是8/xy再问：怎么个约法?再答：y/xyyy约掉剩下1/x

（1）∵xy+x=-1①，xy-y=-2②，∴①-②得x+y=1；（2）先把xy+x=-1，xy-y=-2的值代入代数式，得原式=-x-[2y-1+3x]+2[x+4]=-x-2y+1-3x+2x+8

因为x+2y=0,所以x=-2y原式=(x^2+2xy)/(xy+y^2)=(4y^2-4Y^2)/(-3y^2+y^2)=0/(-2y^2)又因为xy不等于零,所以x、y君不等于零,所以-2y^2亦

(-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*

由已知1x-1y=10，可知：y-x=10xy，∴3y−6xy−3xy+2xy−x=3(y−x)−6xyy−x+2xy=3×10xy−6xy10xy+2xy=24xy12xy＝2．故选B．

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

xy+1/xy>=2√（xy*1/xy）=2(当xy=1/xy即xy=1时取等号)x/y+y/x>=2√（x/y*y/x）=2(当x/y=y/x即x=y取等号)当x=y=1时可以同时满足两项的等号要求

因为xy/（x+y）=1/2所以x+y=2xy原式=3（x+y)-5xy/(-x-y+3xy)=3*2xy-5xy/(-2xy+3xy)=xy/xy=1

xy/(x+y)=2(x+y)/xy=1/21/x+1/y=1/2(3x-xy+3y)/(-x+3xy-y)=(3/y-1+3/x)/(-1/y+3-1/x)=[3(1/x+1/y)-1]/[(3-(

解x²y-xy²=xy(x-y)=5×1=5a²b+ab²=1∴ab(a+b)=1∵a+b=3∴3ab=1∴ab=1/3再问：a(a-b)-a+b=?再答：=a

(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

我没看错题吧?2x的平方+3（xy+y的平方）=2X的平方+xy+3y的平方+2xy=16!?a/a的绝对值可取+1或-1,b/b的绝对值也可取+1或-1所以答案为2,0或-2

(x²+xy)＋(xy+y²)=x²+xy＋xy+y²=x²+2xy+y²=2-1=1