题目内容
(1)先化简,再求值:(x-1-| 8 |
| x+1 |
| x2+2x-3 |
| x2-1 |
| 2 |
(2)若x-3=
| 1 |
| x |
| x2 |
| x4+x2+1 |
分析:(1)首先把括号里因式进行通分,然后把除法运算转化成乘法运算,进行约分化简,最后代值计算.
(2)这道求代数式值的题目,不应考虑通过x-3=
把x的值解出,而是对
进行变形,但要注意在变形过程中应保持代数式的值不变.
(2)这道求代数式值的题目,不应考虑通过x-3=
| 1 |
| x |
| x2 |
| x4+x2+1 |
解答:解:(1)原式=[
-
]•
=
•
=
•
=x-3
当x=3-
时,原式=3-
-3=-
(2)∵x-3=
∴x-
=3
∴原式=
=
=
=
,
| (x+1)(x-1) |
| x+1 |
| 8 |
| x+1 |
| (x+1)(x-1) |
| (x-1)(x+3) |
=
| x2-9 |
| x+1 |
| (x+1)(x-1) |
| (x-1)(x+3) |
=
| (x+3)(x-3) |
| x+1 |
| x+1 |
| x+3 |
=x-3
当x=3-
| 2 |
| 2 |
| 2 |
(2)∵x-3=
| 1 |
| x |
∴x-
| 1 |
| x |
∴原式=
| 1 | ||
x2+1+
|
=
| 1 | ||
(x-
|
=
| 1 |
| 32+3 |
=
| 1 |
| 12 |
点评:第一道题是分式的混合运算需特别注意运算顺序及符号的处理,也需要对通分、分解因式、约分等知识点熟练掌握.第二道题要巧妙地将代数式进行变形,让其含有(x-
),代入求值.
| 1 |
| x |
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