Option 1: because the denominator x? -x+ 1=(x- 1/2)？ - 1/4+ 1 =(x- 1/2)？ +3/4≧3/4 & gt； 0, so its domain is x ∈ r.

Remove the denominator and move the term to get (y-2)x? -(y-2)x+y-3=0

∵x∈R, ∴ Its discriminant δ = (y-2)? -4(y-2)(y-3)=(y-2)[(y-2)-4(y-3)]=(y-2)(-3y+ 10)=-3(y- 10/3)(y-2)⊙0

That is, (y- 10/3)(y-2)≦0, so the range is 2.

[Note y ≠ 2; If y=2, there is 2(x? -x+ 1)=2x？ -2x+3, and thus get 2=3, which is of course ridiculous! So when you use this method, you should leave it.

It means individual extreme cases. ]

Solution 2: y=(2x? -2x+3)/(x？ -x+ 1)=2+ 1/(x？ -x+ 1)

Let y ′ =-(2x-1)/(x? -x+ 1)？ =0, stagnation point x =1/2; When x

x→ ∞limy=x→ ∞lim[2+ 1/(x？ -x+ 1)]= 2； Therefore, the range is 2.

Solution 3: y=(2x? -2x+3)/(x？ -x+ 1)=2+ 1/(x？ -x+ 1)= 2+ 1/[(x- 1/2)？ +3/4]≦2+ 1/(3/4)=2+4/3= 10/3

So there is a maximum10/3; There is no minimum value, but there is a concept of limit (opinion 2), 2