若1/x+1/y=2,则分式(2x+3xy+2y)/(x-xy+y)的值是________。

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第1个回答  2009-07-13
分式(2x+3xy+2y)/(x-xy+y)的分子和分母同时除以xy得:
(2/y +3+ 2/x)/(1/y -1+ 1/x)
=[2(1/x + 1/y)+3]/[(1/x + 1/y)-1]
=[2*2+3]/[2-1]
=7
第2个回答  2009-07-13
(x+y)/xy=2
x+y=2xy
(2x+3xy+2y)/(x-xy+y)
=(4xy+3xy)/(2xy-xy)
=7xy/xy
=7本回答被提问者采纳

若1\/x+1\/y=2,则分式(2x+3xy+2y)\/(x+2xy+y)的值为
第二步,分子分母同时除以xy

若1\/x+1\/y=2 则分式2x+3xy+2y\/x-y+y的值是
解:1\/x+1\/y=2 (x+y)\/(xy)=2 x+y=2xy (2x+3xy+2y)\/(x-xy+y)=[2(x+y)+3xy]\/[(x+y)-xy]=(2·2xy+3xy)\/(2xy-xy)=7xy\/(xy)=7 (2x+3xy+2y)\/(x-xy+y)的值为7。总结:解题思路:①先求出x+y、xy的关系式。②再在所求分式中构造x+y、xy,从而求得分式的值。

已知1\/x+1\/y=2,则2x-3xy+2y\/x+2xy+y的值为?
(2x-3xy+2y)\/(x+2xy+y)...分子分母同时除以xy =(2\/y-3+2\/x)\/(1\/y+2+1\/x)=(4-3)\/(2+2)=1\/4 │a│\/(a^2-a)=1\/(1-a)因为当a>0时,│a│\/(a^2-a)=1\/(a-1)≠1\/(1-a)所以a<0,此时│a│\/(a^2-a)=-a\/(a^2-a)=-1\/(a-1)=1\/(1-a)

若x分之1+y分之1=2,则分式x+2xy+y分之2x+3xy+2y的值为?
1\/x +1\/y=2 则通分 x+y\/xy=2 则 x+y=2xy 所以所求分式 2x+3xy+2y \/ x+2xy+y =[2(x+y)+3xy] \/ [(x+y)+2xy]=7xy\/4xy =7\/4 不明白欢迎追问

若1\/x+1\/y=2,则(2x-xy+2y)÷(3x+5xy+3y)
1\/x+1\/y=2 y+x=2xy (2x-xy+2y)=2(x+y)-xy=3xy (3x+5xy+3y)=3(x+y)+5xy=11xy (2x-xy+2y)÷(3x+5xy+3y)=3xy\/11xy=3\/11

若1\/x+1\/y=2,则(2x-xy+2y)÷(3x+5xy+3y)
1\/x+1\/y=2 y+x=2xy (2x-xy+2y)=2(x+y)-xy=3xy (3x+5xy+3y)=3(x+y)+5xy=11xy (2x-xy+2y)÷(3x+5xy+3y)=3xy\/11xy=3\/11

已知1\/x+1\/y=3,则分式2x+3xy+2y\/x-2xy+y的值为
1\/x+1\/y=(x+y)\/xy=3 即 x+y=3xy (2x+3xy+2y)\/(x-2xy+y)=(2*3xy+3xy)\/(3xy-2xy)=9xy\/xy=9

如果1\/x+1\/y=2求3x+2xy+3y\/2x-3xy+2y的值
3x+2xy+3y\/2x-3xy+2y=(3\/y+2+3\/x)\/(2\/y+2\/x-3)=8\/(4-3)=8

有几道数学题:当1\/x+1\/2y=3时,求分式(2x+3xy+4y)\/(x-3xy+2y)的值_百 ...
1\/x+1\/2y=3 两边乘以2xy得 x+2y=6xy 所以2x+3xy+4y=15xy x-3xy+2y=3xy 所以(2x+3xy+4y)\/(x-3xy+2y)=5 呵呵 给分吧

1\/X +1\/Y=3,则(2X+3XY-2Y)\/(X-2XY-Y)=多少?
解:由已知1\/X + 1\/Y = 3 => (X + Y)\/(XY)= 3 => X + Y = 3XY => XY = (X + Y)\/3,所以原式 = (2X + 3XY –2Y)\/(X –2XY –Y)= (2X + X + Y –2Y)\/[X –2*(X + Y)\/3 –Y]= (3X –Y)\/(X\/3 –5Y\/3)= 3(3X –Y)(X –5Y),(无法再...

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