题目内容
方程x3-
x2-(2
+1)x+3+
=0的三个根分别是
______．
| 3 |
| 3 |
| 3 |
原方程变形得：
x3-(
1) x2+x2-(2
+1) x+3+
=0
x2[x-(
+1)] +[x-(
1)] (x-
)=0
[x-(
)][x2+x-
]=0
即x-(
1) =0或x2+x-
=0
解得：x1=
+1，x2，3=
．
x3-(
| 3+ |
| 3 |
| 3 |
x2[x-(
| 3 |
| 3+ |
| 3 |
[x-(
| 3+1 |
| 3 |
即x-(
| 3+ |
| 3 |
解得：x1=
| 3 |
-1±
| ||||
| 2 |
练习册系列答案
相关题目