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∵(a+b)2≥0或(a-b)2≥0,∴-(a2+b2)≤2ab≤a2+b2,∵4≤a2+b2≤9,進而可得-9≤2ab≤4,解可得,-92≤ab≤2,∴-2≤-ab≤92,∴-2+4≤a2-ab+b2≤92+9,即2≤a2-ab+b2≤272∴所求的最大值與最小值之和是:2+27…
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