已知(a+2)²+|b+3|=0,求3a²b-[2a²b-(2ab-a²b)-4a²]-3ab的值
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已知(a+2)²+|b+3|=0,求3a²b-[2a²b-(2ab-a²b)-4a²]-3ab的值
已知(a+2)²+|b+3|=0,求3a²b-[2a²b-(2ab-a²b)-4a²]-3ab的值
已知(a+2)²+|b+3|=0,求3a²b-[2a²b-(2ab-a²b)-4a²]-3ab的值
(a+2)²+|b+3|=0
则:a=-2,b=-3
原式=4a²-ab
把a=-2,b=-3代入得:
4a²-ab=10
所以,原式=10
∵(a+2)²+|b+3|=0
∴a+2=0
b+3=0
即:a=-2,b=-3
∴3a²b-[2a²b-(2ab-a²b)-4a²]-3ab
=3a²b-2a²b+2ab-a²b+4a²-3ab
=-ab
=-(-2)×(-3)
=-6