已知x y=2,x-y=1求分式

1/x+1/y=3∴x+y=3xy3x-2xy+3y/x+xy+y=[3(x+y)-2xy]/[(x+y)+xy]=(9xy-2xy)/(3xy+xy)=7/4

代入方法：分母：3x-3xy+3y=3(x+y)-3xy=3•(3xy)-3xy=9xy-3xy=6xy分子：x+xy+y=x+y+xy=3xy+xy=4xy所以原式=6xy/4xy=3/

已知1/x-1/y=3,应该是1/x+1/y=3,[按原式得不出结果!]如此,则3xy＝x+y（3x-xy+3y）/（x+xy+y）＝2

x+2y=0x=-2y2xy+y2/x2-xy=-4y^2+y^2/4y^2+2y^2=-5/6

2x-3y=0所以y=2x/3代入(x^2+xy-y^2)/(2x^2+xy-y^2)得到[x^2+x*2x/3-(2x/3)^2]/[2x^2+x*2x/3-(2x/3)^2]=(x^2+2x^2/

1）由1/x-1/y=5,得(y-x)/xy=5,所以y-x=5xy,即x-y=-5xy(3x+5xy-3y)/(x-3xy-y)=[3(x-y)+5xy]/[(x-y)-3xy]=(-15xy+5x

再问：化简：（1）【(a+2/a^2-2a）+(8/4-a^2)】除以a-2/a（2）x-3/x^2-1+1/1-x（3）【a-（2a-1/a）除以1-a^2/a^2+a】再答：嗯嗯~~等我一会

首先化简通分1/x+1/y=5得：（x+y)/xy=5即有x+y=5xy将要求的分式化简：3x+xy+3y/x-2xy+y=3x+3y+xy/x+y-2xy（根据加法交换律）=3（x+y)+xy/(x

x=-2y(2xy+y^2)/(x^2-xy)=(-4y^2+y^2)/(4y^2+2y^2)=-3y^2/(6y^2)=-1/2再问：我算出来也是-1/2，可答案说是-15/28

由1/x-1/y=3两边同乘以xy得y-x=3xy即x-y=-3xy所以(3x+3xy-3y)/(x-2xy-y)＝3(x-y+xy)/(x-y-2xy)=3(-3xy+xy)/(-3xy-2xy)=

因为:1/x-1/y=3(y-x)/xy=3x-y=-3xy（1）又因为：(2x+3xy-2y)/(x-2xy-y)=[2（x-y）+3xy]/[(x-y)-2xy]代入（1）得：（-6xy+3xy）

1/x-1/y=3（y-x）/xy＝3y-x＝3xy（4x-5xy－4y）/（3x-2xy-3y）＝（4（x－y）-5xy）/（3（x－y）－2xy）＝(4*-3xy-5xy)/(3*-3xy－2xy

1/x+1/y=2y+x=2xyx+y=2xy[x+y+2xy+2(y平方)]/(2xy+y平方)=(2xy+2xy+2y^2)/(2xy+y^2)=(4xy+2y^2)/(2xy+y^2)=2(2x

2x=-yy=-2xxy/(x^2-y^2)=x*(-2x)/[x^2-(-2x)^2]=-2x^2/[x^2+4x^2]=-2x^2/5x^2=-2/5

x^2+y^2-6x-8y+25=（x-3)^2+(y-4)^2=0x=3,y=41/xy=1/12

将1/x-1/y=5通分,得y-x/xy=5,两边同时乘以xy,得y-x=5xy,带入-x+xy+y/2x+7xy-2y,得5xy+xy/-2*5xy+7xy,合并同类项,得6xy/-3xy=-2即,

由1/x-1/y=2001得：y-x/xy=2001所以：xy/y-x=1/2001xy/x-y=-1/2001x-xy-y/(x-y)=x-y-xy/(x-y)=1-xy/(x-y)原式=1+1/2

解题方法：分子、分母同时除一个xy(y+xy-x)/(2x+7xy-2y)=[(y+xy-x)/xy]/[(2x+7xy-2y)/xy]=(1/x+1-1/y)/(2/y+7-2/x)=[(1/x-1

(y+xy-x)/(2x+7xy-2y)(分子、分母同除一个xy)=(1/x+1-1/y)/(2/y+7-2/x)=6/(-10+7)=-2

∵x-y/xy=3∴x-y=3xy2x+3xy-2y/x-2xy-y=[2(x-y)+3xy]/[(x-y)-2xy]=(6xy+3xy)/(3xy-2xy)=9xy/(xy)=9