2a²+3ab+b²-a²-ab-2b²,其中a²-b²=2,ab=-3
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2a²+3ab+b²-a²-ab-2b²,其中a²-b²=2,ab=-3
2a²+3ab+b²-a²-ab-2b²,其中a²-b²=2,ab=-3
2a²+3ab+b²-a²-ab-2b²,其中a²-b²=2,ab=-3
=2(a2-b2)-(a2-b2) 2ab
=2x2-2 2x-3=-2
原式合并同类项等于a方+2ab-b方,所以应该等于-4
a²+2ab-b²=2-3=-1
原式=a²+2ab-b² 由已知条件可知 答案为-4计算过程可以说下么