Question

用配方法证明-10x方+7x-4的值恒小于0

Answer 1

原式=-10(x²-7x/10)-4
=-10(x²-7x/10+49/200-49/200)-4
=-10(x²-7x/10+49/200)+49/20-4
=-10(x-7/20)²-31/80≤-31/80<0
所以恒小于0

Answer 2

-10x方+7x-4
=-10(x²-7x/10)-4
=-10(x-7/20)²-4+49/40
=-10(x-7/20)²-111/40;
∵(x-7/20)²≥0恒成立
∴-10x方+7x-4≤-111/40＜0恒成立；

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Answer 3

-10x²+7x-4
=-10(x²-0.7x+0.4)
=-10[(x-0.35)²-0.35²+0.4]
=-10[(x-0.35)²+0.2275]
∵(x-0.35)²+0.2275＞0
∴-10[(x-0.35)²+0.2275]＜0

Answer 4