y=(3x2+3x-1)/(x2+x-1) x属于R求值域

y=(3x²+3x-3+2)/(x²+x-1)
=(3x²+3x-3)/(x²+x-1)+2/(x²+x-1)
=3+2/(x²+x-1)
x²+x-1=(x+1/2)²-5/4>=-5/4
所以1/(x²+x-1)0
2/(x²+x-1)0
3+2/(x²+x-1)3
所以值域(-∞,7/5]∪(3,+∞)