(x-2y)^2+|3x-1|=0
那么只有两项同时为0
即x-2y=0且3x-1=0
所以x=1/3,y=1/6
(24x^2y-12xy^2)/[(3x+y)^2-(3x-y)^2]
=12xy(2x-y)/[(3x+y+3x-y)(3x+y-3x+y)]
=12xy(2x-y)/(12xy)
=2x-y
=2/3-1/6
=4/6-1/6
=1/2