2x+1的绝对值+(3y-1)的二次方=0,则x的二次方-y=

问题描述:

2x+1的绝对值+(3y-1)的二次方=0,则x的二次方-y=

2x+1的绝对值+(3y-1)的二次方=0
∴2x+1=0, 3y-1=0
x=-1/2, y=1/3
x^2-y=1/4 -1/3=-1/12

| 2x+1 | >=0 对吧
(3y -1 )^2 >=0 对吧
所以 2x+1 =0 x =1/2
3y -1=0 y=1/3
x^2 -y =1/4 -1/3 =-1/12

根据题意得:
2x+1=0
3y-1=0
解的:
x=-1/2
y=1/3
x²-y
=(-1/2)²-1/3
=1/4-1/3
=-1/12

2x+1的绝对值+(3y-1)的二次方=0
则:2x+1=0, 3y-1=0
x=-1/2, y=1/3
x^2-y=1/4 -1/3=-1/12