题目内容
(1)计算:(
-
)2+2
×3
(2)解方程:x2-2x-3=0.
| 2 |
| 3 |
|
| 2 |
(2)解方程:x2-2x-3=0.
(1)原式=2-2
+3+
×3
=5-2
+2
=5
(2)移项,得:x2-2x=3
配方x2-2x+12=3+12
(x-1)2=4
则x-1=±2
解得:x1=3,x2=-1
| 6 |
2
| ||
| 3 |
| 2 |
=5-2
| 6 |
| 6 |
=5
(2)移项,得:x2-2x=3
配方x2-2x+12=3+12
(x-1)2=4
则x-1=±2
解得:x1=3,x2=-1
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