题目内容
(1)计算:
(2
+4
-3
)
(2)分解因式:(x2-2x)2+2(x2-2x)+1
(3)先化简,再求值:
÷(x-
),其中x=
+1.
| ||
| 2 |
| 12 |
|
| 48 |
(2)分解因式:(x2-2x)2+2(x2-2x)+1
(3)先化简,再求值:
| x-1 |
| x |
| 2x-1 |
| x |
| 5 |
(1)原式=
×2
+
×4
-
×3
=2
+2×
-
×4
=2
+1-6
=-1-4
;
(2)原式=(x2-2x+1)2
=[(x-1)2]2
=(x-1)4;
(3)
÷(x-
)
=
÷
=
•
=
,
当x=
+1时,原式=
=
.
| ||
| 2 |
| 12 |
| ||
| 2 |
|
| ||
| 2 |
| 48 |
=2
| 6 |
| 1 |
| 2 |
| 3 |
| 2 |
| 6 |
=2
| 6 |
| 6 |
=-1-4
| 6 |
(2)原式=(x2-2x+1)2
=[(x-1)2]2
=(x-1)4;
(3)
| x-1 |
| x |
| 2x-1 |
| x |
=
| x-1 |
| x |
| x2-2x+1 |
| x |
=
| x-1 |
| x |
| x |
| (x-1)2 |
=
| 1 |
| x-1 |
当x=
| 5 |
| 1 | ||
|
| ||
| 5 |
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