A=a2-4ab+4b2+b2+2b++1+99
=(a-2b)2+(b+1)2+99,
∵(a-2b)2≥0,(b+1)2≥0,
∴A≥99,
∴A的最小值为99.
若A=a2+5b2-4ab+2b+100,求A的最小值.
若A=a2+5b2-4ab+2b+100,求A的最小值.
数学人气:136 ℃时间:2019-10-19 10:07:34
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