因为|a-b+3|≥0,(2a+b)²≥0
所以a-b+3=0 ①
2a+b=0 ②
由①+②得3a+3=0,a=-1
将a=-1代入②得-2+b=0,b=2
所以2a²b(2ab+1)-a²(-2ab)²
=4a³b²+2a²b-4(a^4)b²
=2a²b(2ab+1-2a²b)
=2×1×2×[2×(-1)×2+1-2×1×2]
=4×(-8)
=-32
因为|a-b+3|≥0,(2a+b)²≥0
所以a-b+3=0 ①
2a+b=0 ②
由①+②得3a+3=0,a=-1
将a=-1代入②得-2+b=0,b=2
所以2a²b(2ab+1)-a²(-2ab)²
=4a³b²+2a²b-4(a^4)b²
=2a²b(2ab+1-2a²b)
=2×1×2×[2×(-1)×2+1-2×1×2]
=4×(-8)
=-32