Question

y=x的平方-x/x的平方-x+1的值域

Answer 1

y = (x²-x)/(x²-x+1) = (x²-x+1-1)/(x²-x+1) = 1 - 1/(x²-x+1) = 1 - 1/{(x-1/2)²+3/4}
∵(x-1/2)² ≥ 0
∴(x-1/2)²+3/4 ≥ 3/4
∴0 ＜ 1/{(x-1/2)²+3/4} ≤ 4/3
∴-4/3 ≤ - 1/{(x-1/2)²+3/4} ＜ 0
∴1-4/3 ≤ 1 - 1/{(x-1/2)²+3/4} ＜ 1+0
即：-1/3 ≤ 1 - 1/{(x-1/2)²+3/4} ＜ 1
值域【-1/3，1)

Answer 2

y=(x^2-x)/(x^2-x+1),
∴y(x^2-x+1)=x^2-x,
整理得(y-1)x^2+(1-y)x+y=0,①
x,y是实数，y=1时①不成立，
∴y≠1，且(1-y)^2-4y(y-1)=(1-y)(1-y+4y)=(1-y)(3y+1)>=0，
∴-1/3<=y<1,为所求.