Question

已知2x-y+1的绝对值+ （3x+2分之3y）平方=0，求代数式x+y分之y的平方除以（x-y分之x-1）（x-x-y分之x平方

Answer 1

2x-y+1的绝对值+ （3x+2分之3y）平方=0
则2x-y+1=0
3x+3y/2=0，
则2x-y=-1①
2x+y=0②
①+②，得4x=-1，则x=-1/4
①-②，得-2y=-1，则y=-1/2
x+y分之y的平方除以（x-y分之x-1）（x-x-y分之x平方）
=x+y分之y的平方除以（x-y分之y）（x-y分之-xy）
=-x³/(x+y)
=(1/4)³/(-1/4-1/2)
=-1/48

Answer 2

解：
因为|2x-y+1|+(3x+3y/2)^2=0，则
2x-y+1=0
3x+3y/2=0，即
2x-y+1=0
2x+y=0，则
x+y=-x
4x+1=0
x=-1/4
y=1/2

[y^2/(x+y)]/[x/(x-y)-1)][(x-x^2/(x-y)]
=[y^2/(x+y)] [(x-x^2/(x-y)]/ [x/(x-y)-1)]
=[y^2/(x+y)] [(-xy/(x-y)]/ [y/(x-y)]
=[y^2/(-x)][-x]
=y^2
=1/4