问题
解答题
计算: (1)(2
(2)lg52+
|
答案
(1)(2
)-1 4
-3(1 2
-1)-1+π02
=[(
)2]-3 2
-3×1 2
+11
-12
=(
)-1-3×(3 2
+1)+12
=
-32 3
-3+12
=-
-34 3
.2
(2)lg52+
lg8+lg5•lg20+(lg2)22 3
=lg25+lg(23)
+lg5•lg(22×5)+(lg2)22 3
=lg25+lg4+lg5(2lg2+lg5)+(lg2)2
=lg102+2lg5•lg2+(lg5)2+(lg2)2
=2+(lg5+lg2)2
=2+1
=3.