题目内容
| x2-1 |
| x2-2x+1 |
| x2-2x |
| x-2 |
| 1 |
| 2 |
(2)作图题,在AC上找一点D,使得AB2=AD•AC.
分析:(1)通常做法是先把除法统一为乘法,再算加法,把代数式化简,然后再代入求值;
(2)作∠ADB=∠ABC,那么△ABD∽△ACB,∴AB2=AD•AC.
(2)作∠ADB=∠ABC,那么△ABD∽△ACB,∴AB2=AD•AC.
解答:解:(1)
+
÷x
=
+
•
=
+1
=
当x=
时,原式=
=-2;
(2)作法:过点B作∠ADB=∠ABC与AC相交于点D,则点D就是所求的点.

| x2-1 |
| x2-2x+1 |
| x2-2x |
| x-2 |
=
| (x+1)(x-1) |
| (x-1)2 |
| x(x-2) |
| x-2 |
| 1 |
| x |
=
| x+1 |
| x-1 |
=
| 2x |
| x-1 |
当x=
| 1 |
| 2 |
2×
| ||
|
(2)作法:过点B作∠ADB=∠ABC与AC相交于点D,则点D就是所求的点.
点评:主要考查了分式的化简求值和基本作图.分子、分母能因式分解的先因式分解;除法要统一为乘法运算.尺规作图中要掌握画一个角等于已知角的作图方法.
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