因为(a-2)²≥0
√(b+1/2)≥0
所以(a-2)²+√(b+1/2)≥0
要使(a-2)²+√(b+1/2) ≤0
只能(a-2)²+√(b+1/2) =0
得a-2=0
b+1/2=0
解得a=2
b=-1/2
-2(3b²-2a²)+3(ab+2b²-a²)
=-6b²+4a²+3ab+6b²-3a²
=a²+3ab
=2²+3×2×(-1/2)
=4-3
=1
答案:1
因为(a-2)²≥0
√(b+1/2)≥0
所以(a-2)²+√(b+1/2)≥0
要使(a-2)²+√(b+1/2) ≤0
只能(a-2)²+√(b+1/2) =0
得a-2=0
b+1/2=0
解得a=2
b=-1/2
-2(3b²-2a²)+3(ab+2b²-a²)
=-6b²+4a²+3ab+6b²-3a²
=a²+3ab
=2²+3×2×(-1/2)
=4-3
=1
答案:1