龙之谷新区开服时间表:初三的二道数学计算题,本人才疏学浅求助各位高手帮忙解答

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1.已知1/x-1/y=3,求分式2x+3xy-2y/(x-2xy-y)的值

2.已知x/y+z=y/x+z=z/x+y,求x/y+z的值

1.已知1/x-1/y=3,求分式(2x+3xy-2y)/(x-2xy-y)的值
【解】(2x+3xy-2y)/(x-2xy-y)
=(2/y-2/x+3)/(1/y-1/x-2)
=(-6+3)/(-3-2)
=3/5

2.已知x/(y+z)=y/(x+z)=z/(x+y),求x/(y+z)的值
【解】设x/(y+z)=u
u(y+z)=x
u(x+z)=y
u(x+y)=z

三式相加:2u(x+y+z)=x+y+z

若x+y+z=0,显然x/(y+z)=-1

若x+y+z≠0,显然x/(y+z)=u=1/2

1 分式上下都除以XY代入已知
得值为0.6
2 当x=y=z时已知条件相符
故很难求出具体的值

1、上下都除以xy,得:(2/y+3-2/x)/(1/y-2-1/x)=(-2(1/x-1/y)+3)/(-(1/x-1/y)-2)=-3/-5=3/5