已知(a+2)^2+|b+3|=0,求3a^2b-[2a^2b-(2ab-a^2b)-4a^2]-ab的值

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已知(a+2)^2+|b+3|=0,求3a^2b-[2a^2b-(2ab-a^2b)-4a^2]-ab的值
已知(a+2)^2+|b+3|=0,求3a^2b-[2a^2b-(2ab-a^2b)-4a^2]-ab的值

已知(a+2)^2+|b+3|=0,求3a^2b-[2a^2b-(2ab-a^2b)-4a^2]-ab的值
(a+2)^2+|b+3|=0
则:a+2=0,b+3=0
得:a=-2,b=-3
原式=3a²b-(2a²b-2ab+a²b-4a²)-ab
=3a²b-3a²b+2ab+4a²-ab
=4a²+ab
把a=-2,b=-3代入得:
原式=22

(a+2)^2+|b+3|=0
∴﹛a+2=0
b+3=0
∴a=-2, b=-3
3a^2b-[2a^2b-(2ab-a^2b)-4a^2]-ab
=3a²b-(2a²b-2ab+a²b-4a²)-ab
=3a²b-3a²b+2ab+4a²-ab
=ab+4a²
=-2×(-3)+4×(-2)²
=22

已知(a+2)^2+|b+3|=0,
a+2=0,b+3=0
a=-2,b=-3
所以
3a^2b-[2a^2b-(2ab-a^2b)-4a^2]-ab
=3a^2b-2a^2b+2ab-a^2b+4a^2-ab
=a^2b-a^2b+4a^2+(2ab-ab)
=4a^2+ab
=4×(-2)^2+(-2)×(-3)
=16+6
=22

∵a+2)^2+|b+3|=0
∴a+2=0
b+3=0
∴a=-2
b=-3
3a^2b-[2a^2b-(2ab-a^2b)-4a^2]-ab
=3a²b-2a²b+2ab-a²b+4a²-ab
=ab+4a²
当a=-2,b=-3时
原式=6+16=22

由于平方和和绝对值都是 大于等于0的数,因此每一项都等于0,,,,a=-2,,,,b=-3,,,,,带入可求解