当x变化时,分式3x2+6x+5⼀0.5x2+x+1的最小值为多少

2024-11-26 10:18:11
推荐回答(2个)
回答1:

分子分母乘2
原式=(6x²+12x+10)/(x²+2x+2)
=(6x²+12x+12-2)/(x²+2x+2)
=[6(x²+2x+2)-2]/(x²+2x+2)
=6(x²+2x+2)/(x²+2x+2)-2/(x²+2x+2)
=6-2/(x²+2x+2)

x²+2x+2
=x²+2x+1+1
=(x+1)²+1≥1
所以0<1/(x²+2x+2)≤1
-2≤-2/(x²+2x+2)<0
6-2≤6-2/(x²+2x+2)<6+0
4≤6-2/(x²+2x+2)<6
所以最小值=4

回答2:

(3x^2+6x+5)/(0.5x^2+x+1)=
(6(0.5x^2+x+1)-1)/(0.5x^2+x+1)
=6-1/(0.5x^2+x+1)=6-1/(0.5(x+1)^2+0.5)
0.5(x+1)^2+0.5的最小值为0.5
所以原分式最小值为6-1/0.5=4