y=(2x-1)/(x的平方-x+1) 求它的值域

y=(2x-1)/(x的平方-x+1) 求它的值域
y=(2x-1)/(x的平方-x+1) 求它的值域

y=(2x-1)/(x的平方-x+1) 求它的值域
(x^2-x+1)y=2x-1
yx^2-(y+2)x+(y+1)=0
要这个关于x的方程有解必须
Δ=(y+2)^2-4y(y+1)>=0
-2/√3

y=(2x-1)/(x^2-x+1)
=(2x-1)/[(x-1/2)^2+3/4]
=(2x-1)/[(1/4)(2x-1)^2+3/4]
=4(2x-1)/[(2x-1)^2+3]
====>X∈R
====>Y∈R

y=[2（x-1/2）]/[(x-1/2)^2+3/4]
=2/[(x-1/2)+3/4*(x-1/2)]
(x-1/2)+3/4*(x-1/2)>=根号3
所以y<=2/根号3

y=[2(x-1/2)]/[(x-1/2)^2+3/4] 上下同时除以(x-1/2)
得到y=2/[(x-1/2)+3/4(x-1/2)]
对分母利用a+b>=2根号ab，得到分母>=根号3
多以y<=（2倍根号3）/3