Bài 44 trang 15 SBT toán 7 tập 1

Đề bài

Tính:

\(a)\, {25^3}:{5^2};\)

\(b)\,\displaystyle {\left( {{3 \over 7}} \right)^{21}}:{\left( {{9 \over {49}}} \right)^6};\)

\(c)\, \displaystyle 3 - {\left( { - {6 \over 7}} \right)^0} + {\left( {{1 \over 2}} \right)^2}:2\)

Lời giải

\(a)\,{25^3}:{5^2} = {25^3}:25 = {25^2} = 625\) 

\(\displaystyle b)\,{\left( {{3 \over 7}} \right)^{21}}:{\left( {{9 \over {49}}} \right)^6}\)

\(\displaystyle = {\left( {{3 \over 7}} \right)^{21}}:{\left[ {{{\left( {{3 \over 7}} \right)}^2}} \right]^6} \)

\(\displaystyle = {\left( {{3 \over 7}} \right)^{21}}:{\left( {{3 \over 7}} \right)^{12}} \)

\(\displaystyle = {\left( {{3 \over 7}} \right)^9} = {{19683} \over {40353607}}\)

\(c) \,\displaystyle3 - {\left( { - {6 \over 7}} \right)^0} + {\left( {{1 \over 2}} \right)^2}:2\)

\(\displaystyle= 3 - 1 + {\left( {{1 \over 2}} \right)^2}.\frac{1}{2}\)

\(\displaystyle = 2 + {1 \over 8} = 2{1 \over 8}\)


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