|2a-1|+ b?3 =0,∵2a-1≥0, b?3 ≥0,又∵|2a-1|+ b?3 =0,∴2a-1=0,b-3=0,∴a= 1 2 ,b=3,∴[(2a+b)2-(2a-b)(2a+b)-8b]÷2b=[4a2+4ab+b2-4a2+b2-8b]÷2b=(4ab+2b2-8b)÷2b=2a+b-4,把a= 1 2 ,b=3代入得:原式=2× 1 2 +3-4=0.