题目内容
因式分x3-2x2+x=______.
已知x+y=6,xy=-2,则
+
=______.
已知x+y=6,xy=-2,则
| 1 |
| x2 |
| 1 |
| y2 |
x3-2x2+x,
=x(x2-2x+1),
=x(x-1)2;
∵x+y=6,xy=-2,
∴
=
+
=-3,
两边平方得,
+2×
•
+
=9,即
+2×
+
=9
所以,
+
=10.
故答案为:x(x-1)2;10.
=x(x2-2x+1),
=x(x-1)2;
∵x+y=6,xy=-2,
∴
| x+y |
| xy |
| 1 |
| x |
| 1 |
| y |
两边平方得,
| 1 |
| x2 |
| 1 |
| x |
| 1 |
| y |
| 1 |
| y2 |
| 1 |
| x2 |
| 1 |
| xy |
| 1 |
| y2 |
所以,
| 1 |
| x2 |
| 1 |
| y2 |
故答案为:x(x-1)2;10.
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