Nội dung bài giảng
Viết các biểu thức số sau dưới dạng \({{\rm{a}}^n}(a \in Q,n \in N)\):
a) \({9.3^3}.{1 \over {81}}{.3^2}\) b) \({4.2^5}:\left( {{2^3}.{1 \over {16}}} \right)\)
c) \({3^2}{.2^5}.{\left( {{2 \over 3}} \right)^2}\) d) \({\left( {{1 \over 3}} \right)^2}.{1 \over 3}{.9^2}\)
Giải
a) \({9.3^3}.{1 \over {81}}{.3^2} = \left( {{3^2}{{.3}^3}{{.3}^2}} \right).{1 \over {{3^4}}} = {{{3^7}} \over {{3^4}}} = {3^3}\)
b) \({4.2^5}:\left( {{2^3}.{1 \over {16}}} \right) = {2^2}{.2^5}:\left( {{2^3}.{1 \over {{2^4}}}} \right) \)
\(= {2^7}:{1 \over 2} = {2^7}.2 = {2^8}\)
c) \({3^2}{.2^5}.{\left( {{2 \over 3}} \right)^2} = {3^2}{.2^5}.{{{2^2}} \over {{3^2}}} \)
\(= \left( {{3^2}.{1 \over {{3^2}}}} \right).\left( {{2^5}{{.2}^2}} \right) = {1.2^7} = {2^7}\)
d) \({\left( {{1 \over 3}} \right)^2}.{1 \over 3}{.9^2} = \left( {{1 \over {{3^2}}}.{1 \over 3}} \right).{\left( {{3^2}} \right)^2} = {1 \over {{3^3}}}{.3^4} = 3\)