用配方法证明-10x눀+7x-4的值恒小于0

2024-12-16 16:37:34

推荐回答（4个）

回答1：

-10x²+7x-4

=-10(x²-0.7x)-4
=-10(x²-0.7+0.35²)-4+1.225
=-10(x-0.35)²-2.775
-10(x-0.35)²≤0

所以-10x²+7x-4的值恒小于0

回答2：

-10x²+7x-4
=-10(x²-7/10x)-4
=-10(x²-7/10x+49/400)-111/40
=-10(x-7/20)²-111/40
∵-10(x-7/20)²≠0
∵-10(x-7/20)²-111/40＜0
∴-10x²+7x-4的值恒小于0

回答3：

-10x²+7x-4
=-10[x^2-7/10x+(7/20)^2-(7/20)^2+4/10]
=-10[(x-7/20)^2+111/400]
=-10(x-7/20)^2-111/40
因为-10(x-7/20)^2<=0
所以-10(x-7/20)^2-111/40<0

回答4：

-10x²+7x-4=-10(x²-7x/10+4/10)=-10[(x-7/20)²+111/400]=-10(x-7/20)²-111/40<0