题目内容
若x2-3x+1=0,求:
(1)x+
;
(2)x2+
.
(1)x+
1 |
x |
(2)x2+
1 |
x2 |
(1)∵x2-3x+1=0,
∴x≠0,
∴x-3+
=0,
∴x+
=3;
(2)x2+
=(x+
)2-2
=32-2
=7.
∴x≠0,
∴x-3+
1 |
x |
∴x+
1 |
x |
(2)x2+
1 |
x2 |
1 |
x |
=32-2
=7.
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