问题
解答题
计算 (1)(π-2009)0+
(2)
|
答案
(1)原式=1+2
+2-3 3
=(1+2)+(2
-3
)3
=3+
;3
(2)原式=3
-2 3 2
-(1+2
)+1+|1-2
|2
=3
-2 3 2
-1-2
+1+2
-12
=3 2
-1.2
计算 (1)(π-2009)0+
(2)
|
(1)原式=1+2
+2-3 3
=(1+2)+(2
-3
)3
=3+
;3
(2)原式=3
-2 3 2
-(1+2
)+1+|1-2
|2
=3
-2 3 2
-1-2
+1+2
-12
=3 2
-1.2