Bài 1:
\(A = {\left( { - {1 \over 3}} \right)^2}:{1 \over 6} - 2{\left( {{{ - 1} \over 2}} \right)^3} \)
\(\;\;\;\;= {1 \over 9}.6 - 2\left( {{{ - 1} \over 8}} \right)\)
\(\;\;\;\;= {2 \over 3} + {1 \over 4} = {{11} \over {12}}.\)
\(B = \left| { - {3 \over 2} + 1,2} \right| + 1{2 \over 3}:6 \)
\(\;\;\;\;= \left| { - {3 \over 2} + {{12} \over {10}}} \right| + {5 \over 3}:6\)
\(\;\;\;\; = \left| {{{ - 15 + 12} \over {10}}} \right| + {5 \over 3}.{1 \over 6} \)
\(\;\;\;\;= \left| {{{ - 3} \over {10}}} \right| + {5 \over {18}} = {{52} \over {90}} = {{26} \over {45}}.\)
Bài 2:
a) \({3^x} + {3^{x + 2}} = 810\)
\(\eqalign{& \Rightarrow {3^x}\left( {1 + {3^2}} \right) = 810 \cr & \Rightarrow {3^x}.10 = 810 \cr& \Rightarrow {3^x} = 810:10 \cr & \Rightarrow {3^x} = 81 \cr & \Rightarrow {3^x} = {3^4} \cr & \Rightarrow x = 4. \cr} \)
b) \({\left( {x + {{2012} \over {2013}}} \right)^6} = 0.\)
\(\eqalign{ & \Rightarrow x + {{2012} \over {2013}} = 0 \cr & \Rightarrow x = - {{2012} \over {2013}}. \cr} \)
