题目内容
若x2-
x+1=0,则x4+
=
.
| ||
| 2 |
| 1 |
| x4 |
| 89 |
| 16 |
| 89 |
| 16 |
分析:先把x2-
x+1=0两边除以x2可得x+
=
,然后根据完全平方公式对x4+
进行变形得(x2+
)2-2=[(x+
)2-2]2-2,再把x+
=
整体代入进行计算即可.
| ||
| 2 |
| 1 |
| x |
| ||
| 2 |
| 1 |
| x4 |
| 1 |
| x2 |
| 1 |
| x |
| 1 |
| x |
| ||
| 2 |
解答:解:∵x2-
x+1=0,
∴x-
+
=0,即x+
=
,
x4+
=(x2+
)2-2
=[(x+
)2-2]2-2
=[(
)2-2]2-2
=(
)2-2
=
.
故答案为
.
| ||
| 2 |
∴x-
| ||
| 2 |
| 1 |
| x |
| 1 |
| x |
| ||
| 2 |
x4+
| 1 |
| x4 |
| 1 |
| x2 |
=[(x+
| 1 |
| x |
=[(
| ||
| 2 |
=(
| 11 |
| 4 |
=
| 89 |
| 16 |
故答案为
| 89 |
| 16 |
点评:本题考查了完全平方公式:(a±b)2=a2±2ab+b2.也考查了代数式的变形能力以及整体思想的运用.
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