由题,ax²+bx+c≥0恒成立,则有①a=b=0,c＞0与题设a＜b不符,舍去②a＞0,△=b²-4ac≤0则4ac≥b²易知M=(a+2b+4c)/(b-a)=[a·(a+2b+4c)][a·(b-a)]=(a²+2ab+4ac)/(ab-a²)∴M≥(a²+2ab+b
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