Bài 45 trang 15 SBT toán 7 tập 1

Đề bài

Viết các biểu thức số sau dưới dạng \({{\rm{a}}^n}(a \in\mathbb Q,n \in\mathbb N)\):

a) \(\displaystyle {9.3^3}.{1 \over {81}}{.3^2}\) 

b) \(\displaystyle {4.2^5}:\left( {{2^3}.{1 \over {16}}} \right)\)

c) \(\displaystyle {3^2}{.2^5}.{\left( {{2 \over 3}} \right)^2}\) 

d) \(\displaystyle {\left( {{1 \over 3}} \right)^2}.{1 \over 3}{.9^2}\)

Lời giải

a) \(\displaystyle {9.3^3}.{1 \over {81}}{.3^2} = \left( {{3^2}{{.3}^3}{{.3}^2}} \right).{1 \over {{3^4}}}\)\(\displaystyle \,= {{{3^7}} \over {{3^4}}} = {3^3}\)

b) \(\displaystyle {4.2^5}:\left( {{2^3}.{1 \over {16}}} \right) = {2^2}{.2^5}:\left( {{2^3}.{1 \over {{2^4}}}} \right) \)\(\,\displaystyle = {2^7}:{1 \over 2} = {2^7}.2 = {2^8}\) 

c) \(\displaystyle {3^2}{.2^5}.{\left( {{2 \over 3}} \right)^2} = {3^2}{.2^5}.{{{2^2}} \over {{3^2}}} \)

\(\displaystyle = \left( {{3^2}.{1 \over {{3^2}}}} \right).\left( {{2^5}{{.2}^2}} \right) = {1.2^7} = {2^7}\)

d) \(\displaystyle {\left( {{1 \over 3}} \right)^2}.{1 \over 3}{.9^2} = \left( {{1 \over {{3^2}}}.{1 \over 3}} \right).{\left( {{3^2}} \right)^2}\)\(\,\displaystyle = {1 \over {{3^3}}}{.3^4} = 3^1\).


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