求证:-1≤x-1/x^2-x+1≤1/3
答案:3 悬赏:70
解决时间 2021-11-05 19:00
- 提问者网友：轮囘Li巡影
- 2021-11-05 12:44
求证:-1≤x-1/x^2-x+1≤1/3
最佳答案
- 二级知识专家网友：逃夭
- 2021-11-05 14:13
解：∵ (x-2)²≥0
∴(x²-4x+4)≥0
又∵x²-x+1=(x-1/2)²+3/4>0
∴(x²-4x+4)/(x²-x+1)≥0
(x²-x+1-3x+3)/(x²-x+1)≥0
1/3-(x-1)/(x²-x+1)≥0
(x-1)/(x²-x+1)-1/3≤0
(x-1)/(x²-x+1)≤1/3 ①
∵x²≥0，x²-x+1=(x-1/2)²+3/4>0
∴x²/(x²-x+1)≥0
(x-1+x²-x+1)/(x²-x+1)≥0
(x-1)/(x²-x+1)+1≥0
(x-1)/(x²-x+1)≥ -1 ②
由①②得-1≤x-1/x²-x+1≤1/3。
∴(x²-4x+4)≥0
又∵x²-x+1=(x-1/2)²+3/4>0
∴(x²-4x+4)/(x²-x+1)≥0
(x²-x+1-3x+3)/(x²-x+1)≥0
1/3-(x-1)/(x²-x+1)≥0
(x-1)/(x²-x+1)-1/3≤0
(x-1)/(x²-x+1)≤1/3 ①
∵x²≥0，x²-x+1=(x-1/2)²+3/4>0
∴x²/(x²-x+1)≥0
(x-1+x²-x+1)/(x²-x+1)≥0
(x-1)/(x²-x+1)+1≥0
(x-1)/(x²-x+1)≥ -1 ②
由①②得-1≤x-1/x²-x+1≤1/3。
全部回答
- 1楼网友：污到你湿
- 2021-11-05 16:26
-1≤(x-1)/(x^2-x+1)≤1/3
(x^2-x+1)=(x-1/2)^2+3/4>0
-3(x^2-x+1)≤3(x-1)≤(x^2-x+1)
-3x^2+3x-3≤3x-3≤x^2-x+1
-3x^2≤0≤x^2-4x+4
-3x^2≤0, x^2-4x+4=(x-2)^2>=0
上面的步骤反推，就求证过程。
- 2楼网友：西岸风
- 2021-11-05 15:07
(x-1)(x^2-x+1) =x^3-x^2+x-x^2+x-1 =x^3-2x^2+2x-1
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