题目内容
在实数范围内分解因式:3x2-6x+1=______.
3x2-6x+1
=3(x2-2x+
)
=3[(x-1)2-
]
=3(x-1+
)(x-1-
)
=3(x-
)(x-
).
故答案为3(x-
)(x-
).
=3(x2-2x+
| 1 |
| 3 |
=3[(x-1)2-
| 2 |
| 3 |
=3(x-1+
| ||
| 3 |
| ||
| 3 |
=3(x-
3+
| ||
| 3 |
3-
| ||
| 3 |
故答案为3(x-
3+
| ||
| 3 |
3-
| ||
| 3 |
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