若x y=2 7,则分式-搜答案-拍题作业网

1.选D1/x+1/y=2,通分得x+y/xy=2即x+y=2xy原式=2（x+y）+3xy/x+y+2xy=7xy/4xy=7/42.x+y=3x²+xy+3y=x（x+y）+3y=3x+

1/X+1/Y=2两边同乘以XY,得Y+X=2XY,于是,XY=（X+Y)/2(2X+3XY+2Y)/(X-XY+Y)=7楼上的刚好写反了.

1/x+1/y=2则(2x+3xy+2y)/(x-xy+y)分子分母同除以xy=(2/x+2/y+3)/(1/x+1/y-1)=(4+3)/(2-1)=7

1/x+1/y=2通分(x+y)/xy=2所以x+y=2xy(2x+3xy+2y)/(x-xy+y)=[2(x+y)+3xy]/[(x+y)-xy]把x+y=2xy代入=(2*2xy+3xy)/(2x

(x+y)/xy=2x+y=2xy(2x+3xy+2y)/(x-xy+y)=(4xy+3xy)/(2xy-xy)=7xy/xy=7

1/x-1/y=3(y-x)/xy=3y-x=3xy(2x+3xy-2y)/(3x-2xy-3y)=(3xy-6xy)/(-2xy-9xy)=3/11

即y=-2x所以y²=4x²xy=-2x²所以原式=-2x²/)x²-4x²)=-2x²/(-3x²)=2/3

xy=x-y两边同时除以xy1=1/y-1/xy分之一减x分之一等于1

原式=x−yxy＝1．故选C．

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

∵x/y=2∴x=2y∴(x²-y²)/(xy)=(4y²-y²)/(2y²)=3y²/2y²=3/2

∵x+3y=0，xy≠0，∴x=-3y，x≠0，y≠0，∴x2−3xyxy+y2=x(x−3y)y(x+y)=3y(−3y−3y)y(−3y+y)=−18−2=9．故答案是：9．

应该是（x²-y²）/xy吧3/2

x+2y=0,xy不等于0∴x=-2yx²+2xy/xy+y²=(4y²-4y²)/(-2y²+y²)=0

x^2+xy-2y^2=0(x+2y)(x-y)=0x=-2y或x=yx=-2y原式=(4y^2-6y^2+3y^2)/(4y^2+y^2)=y^2/5y^2=1/5x=y原式=(y^2+3y^2+3

3/5再问：详细过程呢再答：y-x=3xyx-y=-3xy代入得：(-6xy+3xy)/(-3xy-2xy)=-3/-5=3/5

若x/y=2,则分式x²-y²/xy的值=x/y-y/x=2-1/2=3/2

上下同时除以y的平方,得到5/12

（Y－X）÷（XY）＝3可变形为Y-X=3XY（2X＋3XY－2Y）÷（X－2XY－Y）=(2X-2Y+3XY）÷(X-Y-2XY)={2（x-y)+3xy}÷(X-Y-2XY)={2(3xy)+3x

2xy/x+y2x.2y/2x+2y=4xy/2*(x+y)=2xy/x+y不知道对不对