问题
解答题
(1)解方程:x2-2x-1=0; (2)计算:(3
|
答案
(1)移项,得:x2-2x=1,
配方:x2-2x+1=2,
即(x-1)2=2,
则x-1=±
2 |
因而x1=
2 |
2 |
(2)原式=(3
2 |
3 |
2 |
3 |
2 |
3 |
=(3
2 |
3 |
3 |
=-12
6 |
(1)解方程:x2-2x-1=0; (2)计算:(3
|
(1)移项,得:x2-2x=1,
配方:x2-2x+1=2,
即(x-1)2=2,
则x-1=±
2 |
因而x1=
2 |
2 |
(2)原式=(3
2 |
3 |
2 |
3 |
2 |
3 |
=(3
2 |
3 |
3 |
=-12
6 |