问题
解答题
计算: (1)(
(2)解方程2x2+x-1=0. |
答案
(1)原式=3
-35
-22
+52 5
=8
-55
;2
(2)2x2+x-1=0,
分解因式得:(2x-1)(x+1)=0,
2x-1=0,x+1=0,
解得:x1=
,x2=-1.1 2
计算: (1)(
(2)解方程2x2+x-1=0. |
(1)原式=3
-35
-22
+52 5
=8
-55
;2
(2)2x2+x-1=0,
分解因式得:(2x-1)(x+1)=0,
2x-1=0,x+1=0,
解得:x1=
,x2=-1.1 2