x2+x+1 |
x2+1 |
x2+1 |
x2+x+1 |
2 |
3 |
3 |
2 |
设y=
x2+x+1 |
x2+1 |
1 |
y |
2 |
3 |
3 |
2 |
解得y1=
2 |
3 |
3 |
2 |
当
x2+x+1 |
x2+1 |
2 |
3 |
-3±
| ||
2 |
当
x2+x+1 |
x2+1 |
3 |
2 |
经检验,x=
-3±
| ||
2 |
故原方程的根是x=
-3±
| ||
2 |
x2+x+1 |
x2+1 |
2 x2+x+2 |
x2+x+1 |
19 |
6 |
x2+x+1 |
x2+1 |
x2+1 |
x2+x+1 |
2 |
3 |
3 |
2 |
x2+x+1 |
x2+1 |
1 |
y |
2 |
3 |
3 |
2 |
2 |
3 |
3 |
2 |
x2+x+1 |
x2+1 |
2 |
3 |
-3±
| ||
2 |
x2+x+1 |
x2+1 |
3 |
2 |
-3±
| ||
2 |
-3±
| ||
2 |