Question

y=x^2-3x+2/x^2-x+1,判别式法

Answer 1

求值域?

y=(x^2-3x+2)/(x^2-x+1)

(x^2-x+1)y=x^2-3x+2

yx^2-yx+y=x^2-3x+2

(y-1)x^2+(3-y)x+(y-2)=0

当y=1时

0+2x-1=0

x=1/2满足定义域R

当y≠1时

Δ=(3-y)^2-4(y-1)(y-2)>=0

解得

(3-2√3)/3