|3-a|+(b+1)²=0
3-a=0 b+1=0
a=3 b=-1
于是
(2b-a+m)/2-(1/2b-a+m)=2
(-2-3+m)/2-(-1/2-3+m)=2
-5/2+m/2+1/2+3-m=2
m/2-m+1=2
-m/2=1
m=-2
|3-a|+(b+1)²=0
3-a=0 b+1=0
a=3 b=-1
于是
(2b-a+m)/2-(1/2b-a+m)=2
(-2-3+m)/2-(-1/2-3+m)=2
-5/2+m/2+1/2+3-m=2
m/2-m+1=2
-m/2=1
m=-2