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51问答网 > 当x变化时,分式3x눀+6x+5⼀(1⼀2)x눀+x+1的最小值是（），求过程啊，谢谢

当x变化时,分式3x눀+6x+5⼀(1⼀2)x눀+x+1的最小值是（），求过程啊，谢谢

2024-11-26 21:53:42

推荐回答（4个）

回答1：

(3x²+6x+5)/[(1/2)x²+x+1]
=6(x²+2x+2-1/3)/(x²+2x+2)
=6-2/(x²+2x+1+1)
=6-2/[(x+1)²+2]
当x+1=0，即x=-1时(x+1)²+2最小值为2，2/[(x+1)²+2]最大值为1，-2/[(x+1)²+2]最小值为-2
所以原分式的最小值是4

回答2：

当x变化时,分式3x²+6x+5/(1/2)x²+x+1的最小值是（4 ），求过程啊，谢谢

(3x²+6x+5)/(1/2)x²+x+1
=2(3x²+6x+5)/(x²+2x+2)
=(6x²+12x+12-2)/(x²+2x+2)
=6-2/(x²+2x+2)
=6-2/[(x+1)²+1]
[(x+1)²+1]有最小值1
2/[(x+1)²+1]有最大值2
6-2/[(x+1)²+1]有最小值4

回答3：

凑完全平方式
然后就出结果了

回答4：

5/(1/2)x²看不懂

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