已知x-y=-2,求(x^2 y^2)^2

(2x+y)/(x^2-2xy+y^2)X(x-y)=(2x+y)/（x-y）²*（x-y)=(2x+y)/（x-y）x-3y=0,x=3y(2x+y)/（x-y）=（6y+y）/（3y-y

2/x+2/y=根号24(2y+2x)/xy=2√6x+y=xy√6x/y(x-y)-y/x(x-y)=1/(x-y)[(x/y-y/x)]=1/(x-y)[(x²-y²)/xy]

你的题目有点问题我这样做了x/{y(x-y)}-y/{x(x-y)}=（x平方-y平方）/{xy(x-y)}=(x+y)/xy2/x+2/y=2(x+y)/xy=根号24

先化简[(x²+y²)-(x-y)²+2y(x-y)]/4y[(x²+y²)-(x-y)²+2y(x-y)]/4y=[x²+y&s

(x+y)/x=11/88x+8y=11y8x=3yy=8x/3(x-y)/(x+2y)=(x-8x/3)/(x+16x/3)=(-5x/3)/(19x/3)=-5/19

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

这道题实际就是要把x^2+y^2变换成只由x+y和y组成的多项式x^2+y^2=x^2-y^2+2y^2=(x+y)(x-y)+2y^2=(x+y)[(x+y)-2y]+2y^2将式中(x+y)替换为

3x=8yx/y=8/3（1）x+y/y=x/y+1=8/3+1=11/3(2)2x+3y/x-2y分子分母同时除以y得=(2x/y+3)/(x/y-2)=(16/3+3)/(8/3-2)=(25/3

∵2x-y/x+2y=3∴2(2x-y)/x+2y-9(x+2y)/2x-y+3＝2×3-9×1/3+3=6

X/X···这是什么意思···再问：x减y分之x等于x分之x减2y再答：两边同乘x（x-2y）再问：再答：题目是什么····就给个图啊···再问：如图，已知EFGH为矩形，AD⊥Bc于D，交HG于K，

X+Y分之X-Y等于3x=-2yX+Y分之2(x-y)减X+Y分之3X+Y=(-x-3y)/(x+y)=1

因为(x-y)/(x+y)=3,则(x＋y)/(x－y)＝1/3则5(x-y)(x+y)-(x+y)/2(x-y)＝5*3-1/(3*2)=15-1/6=89/6

（x+y)（x-y)-y^2+（x-y)^2-（6x^2y-2xy^2)÷2y=（x+y)（x-y)-y^2+（x-y)^2-2y（3x^2-xy)÷2y=（x+y)（x-y)-y^2+（x-y)^2

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

如果是(x-y)/x=x/x-2y,则x:y=2:3(x-y)/x=（x-2y+2y）/x-2y1-y/x=1+2y/(x-2y)-y/x=2y/(x-2y)x/y=(x-2y)/-2yx/y=-1/

4x=9yx=9/4*y(1)(x+y)/y=[(9/4)y+y]/y=(9/4+1)y/y=9/4+1=13/4(2)(y-x)/2x=[y-(9/4)y]/[2*(9/4)y]=(1-9/4)y/

x+y/y=11/8说明11y=8（x+y）即8x=3y所以x-y/x+2y=3（x-y）/3(x+2y)=（3x-3y）/(3x+6y)=(3x-8x)/(3x+16x)=-5/19

1,-3再问：过程。。。再答：★（x²－2x)+（y²－4y）=5★（x－1）²+（y－2）²=1+4－5★（x－l）²=0,（y－2）²=

原式化简=x＋2y=2011

原式=x-x+x-x+……-x+(2-1+4-3+5-4+……+2008-2007-2009）y=0+(1×1004-2009)y=-1005y=1005/2