y=2x²-2x+3/x²-x+1麻烦用分离常数法做一下,Δ就不用了

Answer 1

y=(2x²-2x+3)/(x²-x+1)
=(2x²-2x+2+1)/(x²-x+1)
=[2(x²-x+1)+1]/(x²-x+1)
=2+ 1/(x²-x+1)
令y=2+1/u
u=x²-x+1=(x-1/2)²+3/4≥3/4
∵y=2+1/u在u∈[3/4,+∞）单调递减
∴当u=3/4时,函数y取得最大值,最大值为y=2+1/(3/4)=2+4/3=10/3
因此原函数的值域为（-∞,10/3]