2x2+2x+3 |
x2+x+1 |
2(x2+x+1)+1 |
x2+x+1 |
1 |
x2+x+1 |
且x2+x+1=(x+
1 |
2 |
3 |
4 |
3 |
4 |
∴0<
1 |
x2+x+1 |
4 |
3 |
∴2<2+
1 |
x2+x+1 |
10 |
3 |
即2<y≤
10 |
3 |
∴函數(shù)y的值域是(2,
10 |
3 |
故答案為:(2,
10 |
3 |
2x2+2x+3 |
x2+x+1 |
2x2+2x+3 |
x2+x+1 |
2(x2+x+1)+1 |
x2+x+1 |
1 |
x2+x+1 |
1 |
2 |
3 |
4 |
3 |
4 |
1 |
x2+x+1 |
4 |
3 |
1 |
x2+x+1 |
10 |
3 |
10 |
3 |
10 |
3 |
10 |
3 |