用配方法证明:-10x눀+7x-4的值恒小于零

2024-12-21 12:28:31
推荐回答(1个)
回答1:

证明:
-10x²+7x-4
=-10[x²-(7/10)x+(2/5)]
=-10[x²-(7/10)x+(7/20)²-(7/20)²+(2/5)]
=-10{[x²-(7/10)x+(7/20)²]+[-(7/20)²+(2/5)]}
=-10[(x-7/20)²+(111/400)]
=-10(x-7/20)²-(111/40)
≤-(111/40)
<0

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