题目内容
(1)计算:9
÷3
×
;
(2)解方程:4x2-3x-1=0(用配方法).
| 45 |
|
| 3 |
| 2 |
2
|
(2)解方程:4x2-3x-1=0(用配方法).
(1)原式=27
÷
×
×
=27
×
×
×
=45
;
(2)4x2-3x-1=0,移项,得4x2-3x=1,
二次项系数化为1,得x2-
x=
,
配方,得x2-
x+(
)2=
+(
)2,
即(x-
)2=
,
解得:x-
=±
,即x=
±
,
则x1=1,x2=
.
| 5 |
3
| ||
| 5 |
| 3 |
| 2 |
2
| ||
| 3 |
| 5 |
| ||
| 3 |
| 3 |
| 2 |
2
| ||
| 3 |
| 6 |
(2)4x2-3x-1=0,移项,得4x2-3x=1,
二次项系数化为1,得x2-
| 3 |
| 4 |
| 1 |
| 4 |
配方,得x2-
| 3 |
| 4 |
| 3 |
| 8 |
| 1 |
| 4 |
| 3 |
| 8 |
即(x-
| 3 |
| 8 |
| 25 |
| 64 |
解得:x-
| 3 |
| 8 |
| 5 |
| 8 |
| 3 |
| 8 |
| 5 |
| 8 |
则x1=1,x2=
| 1 |
| 4 |
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