问题
填空题
计算: (1)
(2)1.10+
|
答案
(1)∵3<π,
∴
=|3-π|=π-3.(3-π)2
故答案为:π-3.
(2)1.10+
-0.5-2+lg25+2lg23 64
=1+4-
+lg1001 (
)21 2
=5-4+2
=3.
故答案为:3.
计算: (1)
(2)1.10+
|
(1)∵3<π,
∴
=|3-π|=π-3.(3-π)2
故答案为:π-3.
(2)1.10+
-0.5-2+lg25+2lg23 64
=1+4-
+lg1001 (
)21 2
=5-4+2
=3.
故答案为:3.