若是这样,则方法如下:
由1/x-1/y=5,得:x、y都不为0.
∴(2x+3xy-2y)/(y+2xy-x)的分子和分母同除以(xy),得:
(2x+3xy-2y)/(y+2xy-x)
=(2/y+3-2/x)/(1/x+2-1/y)
=[3-2(1/x-1/y)]/[2+(1/x-1/y)]
=(3-2×5)/(2+5)
=-1
注:若原题不是我所猜测的那样,则请你补充说明.
若是这样,则方法如下:
由1/x-1/y=5,得:x、y都不为0.
∴(2x+3xy-2y)/(y+2xy-x)的分子和分母同除以(xy),得:
(2x+3xy-2y)/(y+2xy-x)
=(2/y+3-2/x)/(1/x+2-1/y)
=[3-2(1/x-1/y)]/[2+(1/x-1/y)]
=(3-2×5)/(2+5)
=-1
注:若原题不是我所猜测的那样,则请你补充说明.