问题
解答题
计算. (1)(1-π)0-
(2)(
|
答案
(1)原式=1-
×3
+(-8)×3 2 1 16
=1-
-3 2 1 2
=-1;
(2)原式=2-
×2
+3×12 2
=2-1+3
=4.
计算. (1)(1-π)0-
(2)(
|
(1)原式=1-
×3
+(-8)×3 2 1 16
=1-
-3 2 1 2
=-1;
(2)原式=2-
×2
+3×12 2
=2-1+3
=4.