已知x>0,y>0,8x 2y-xy=0
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x2y+xy2-x-y=xy(x+y)-(x+y)=(x+y)(xy-1)∵x+y=-5,xy=7,∴原式=-5×(7-1)=-30.
∵x+y=6,xy=4,∴x2y+xy2=xy(x+y)=4×6=24.故答案为:24.
原式=(x4-xy3)+(y4-x3y)+(3xy2-3x2y)=x(x3-y3)+y(y3-x3)+3xy(y-x)=(x3-y3)(x-y)-3xy(x-y)=(x-y)(x3-y3-3xy)=(
x3+y3-x2y-xy2=(x+y)(x2-xy+y2)-xy(x+y)=(x+y)(x2-2xy+y2)=(x+y)(x2+2xy+y2-4xy)=(x+y)[(x+y)2-4xy]=10×(10
1数1挪到等式左边2变化-13因式分解4推理得出结论答案5连年
把X=3;Y=-2代入即可解
∵xy+x+y+7=0  
那个2是平方吧?可以用^代替原式=x^y+xy^=xy(x+y)=-3*6=-18
是不是求:5x²y-[2x²-(3xy-xy²)-3x²]-2xy²-y²再问:是再答:已知是不是(x+3)²+|x+y+10|=
(xy2-x)dx+(x2y+y)dy=0y(x²+1)dy=-x(y²-1)dxy/(y²-1)dy=-x/(x²+1)dx两边积分得ln|y²-1
|x-2|+(y+3)²=0都是非负式所以分别都=0所以x-2=0y+3=0所以x=2y=-3又因为z是最大的负整数所以z=-1原式=2(x²y+xyz)-3(x²y-x
(x2+z2)(x2+y2)(y2+z2)=(x+y)2-2xy×(x+z)2-2xz×(y+z)2-2yz--之后不清楚了
原式=5xy2-2x2y+3xy2-2x2y=8xy2-4x2y,∵(x-2)2+|y+1|=0,∴x-2=0,y+1=0,即x=2,y=-1,则原式=16+16=32.
(x+2)²+|y-1|=0平方数与绝对值都是非负数两个非负数的和为0,那么这两个数都是0x+2=0y-1=0解得:x=-2,y=1x³+3x²y+3xy²+y
x2y+xy2=xy*(x+y)因为x+y=-(7+xy)又x+y=(9+2xy)\3所以(9+2xy)\3=-(7+xy)3+2xy\3=-7-xy5xy\3=-10解得xy=-6所以x+y=-(7
由题意得(x-2)平方+(y-2)平方+(x-y)平方=0,故x=y=2,故x平方y=8
(x+2)²+|y-1|=0平方数与绝对值都是非负数两个非负数的和为0,那么这两个数都是0x+2=0y-1=0解得:x=-2,y=1x³+3x²y+3xy²+y
由题意得:3C=A+B=8x2y-6xy2-3xy+7xy2-2xy+5x2y=13x2y+xy2-5xy,∴C=13x2y+xy2−5xy3,故:C-A=13x2y+xy2−5xy3-(8x2y-6
x²+4y²-4x+4y+5=0(x-2)²+(2y+1)²=0x-2=0x=22y+1=0y=-1/2x-y=2+1/2=5/2x²y-xy
x²-x=7y²-y=7相减x²-x-y²+y=0(x+y)(x-y)=x-yx-y≠0约分x+y=1x²-x=7y²-y=7相加x&sup