题目内容
已知x+
-3=0,求值:
(1)x2+
(2)x-
.
| 1 |
| x |
(1)x2+
| 1 |
| x2 |
(2)x-
| 1 |
| x |
(1)∵x+
-3=0,
∴x+
=3,
∴x2+
=(x+
)2-2=9-2=7,
即x2+
=7;
(2)由(1)知,x2+
=7,
∴(x-
)2=x2+
-2=7-2=5,
∴x-
=±
.
| 1 |
| x |
∴x+
| 1 |
| x |
∴x2+
| 1 |
| x2 |
| 1 |
| x |
即x2+
| 1 |
| x2 |
(2)由(1)知,x2+
| 1 |
| x2 |
∴(x-
| 1 |
| x |
| 1 |
| x2 |
∴x-
| 1 |
| x |
| 5 |
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