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)/(2xy-xy)
=7xy/xy
=7
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。
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
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
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
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\/11,2,(2x-xy+2y)÷(3x+5xy+3y)=(2\/y-1+2\/x)\/3\/y+5+3\/x)=3\/11,0,
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x分之一+y分之一=2 x-xy+y分之2x+3xy+2y
先算出前面的X分之一+Y分之一，化简得到X+Y=2XY，那么化简后面的x-xy+y分之2x+3xy+2y，得X+Y-XY分之2X-3XY+2Y,然后把X+Y=2XY代入原式得X+Y-XY分之2（X+Y)+3XY,得7。
已知1\/x+1\/y=5,求分式2x-3xy+2y\/x+2xy+y的值
∵1\/x+1\/y=5 ∴(2x-3xy+2y)\/(x+2xy+y)=(2\/y-3+2\/x)\/(1\/y+2+1\/x) ...【分子分母同除以xy】={2(1\/x+1\/y)-3}\/{(1\/x+1\/y)+2} =(2*5-3)\/(5+2)=7\/7 =1
有几道数学题:当1\/x+1\/2y=3时,求分式(2x+3xy+4y)\/(x-3xy+2y)的值
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
