题目内容
(1)先化简,再求值:| x-2 |
| x2-1 |
| 2x+2 |
| x2+2x+1 |
| 1 |
| x-1 |
| 2 |
(2)解方程:x2-3x-1=0.
分析:(1)先把各分子分母因式分解,再把除转化为乘,然后进行约分,最后为同分母加法;把x=
+1代入化简的结果中,分母有理化即可.
(2)把a=1,b=-3,c=-1代入求根公式进行计算即可.
| 2 |
(2)把a=1,b=-3,c=-1代入求根公式进行计算即可.
解答:解:(1)原式=
÷
+
,
=
•
+
,
=
+
,
=
,
=
,
当x=
+1时,
原式=
=
=
=
;
(2)∵a=1,b=-3,c=-1,
∴b2-4ac=(-3)2-4×1×(-1)=13,
∴x=
=
,
∴x1=
,x2=
.
| x-2 |
| (x+1)(x-1) |
| 2(x+1) |
| (x+1)2 |
| 1 |
| x-1 |
=
| x-2 |
| (x+1)(x-1) |
| (x+1)2 |
| 2(x+1) |
| 1 |
| x-1 |
=
| x-2 |
| 2(x-1) |
| 1 |
| x-1 |
=
| x |
| 2(x-1) |
=
| x |
| 2x-2 |
当x=
| 2 |
原式=
| ||
2
|
| ||
2
|
(
| ||||
2
|
2+
| ||
| 4 |
(2)∵a=1,b=-3,c=-1,
∴b2-4ac=(-3)2-4×1×(-1)=13,
∴x=
3±
| ||
| 2×1 |
3±
| ||
| 2 |
∴x1=
3+
| ||
| 2 |
3-
| ||
| 2 |
点评:本题考查了一元二次方程ax2+bx+c=0(a≠0,a,b,c为常数)的求根公式:x=
(b2-4ac≥0).也考查了分式的化简与二次根式的运算.
-b±
| ||
| 2a |
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