题二：求函数y=（2x^2-2x+3)/(x^2-x+1) 的值域。
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y=(2x²-2x+2+1)/(x²-x+1)
=(2x²-2x+2)/(x²-x+1)+1/(x²-x+1)
=2+1/(x²-x+1)
x²-x+1=(x-1/2)²+3/4>=3/4
所以0<1/(x²-x+1)<=4/3
2<2+1/(x²-x+1)<=10/3
值域(2,10/3]
=(2x²-2x+2)/(x²-x+1)+1/(x²-x+1)
=2+1/(x²-x+1)
x²-x+1=(x-1/2)²+3/4>=3/4
所以0<1/(x²-x+1)<=4/3
2<2+1/(x²-x+1)<=10/3
值域(2,10/3]
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