Đề bài

Tính: \(\eqalign{  & a)\,\,C = \left| { - {1 \over 2}} \right| + {\left( { - {1 \over 3}} \right)^2}:\left| { - 2} \right| - {\left( { - {2 \over 3}} \right)^0}  \cr  & b)\,\,D = {{{2^{15}}{{.9}^3}} \over {{6^7}{{.4}^4}}}  \cr  & c)\,\,E = \left[ {{{\left( { - {1 \over 3}} \right)}^2}.{{27} \over 7} + \sqrt {{4 \over {49}}}  - 3} \right]:{4 \over 7} \cr} \)

Lời giải chi tiết

\(\eqalign{  & a)C = \left| { - {1 \over 2}} \right| + {\left( { - {1 \over 3}} \right)^2}:\left| { - 2} \right| - {\left( { - {2 \over 3}} \right)^0} = {1 \over 2} + {1 \over 9}:2 - 1  \cr  &  = {1 \over 2} + {1 \over {18}} - 1 = {9 \over {18}} + {1 \over {18}} - {{18} \over {18}} = {{ - 8} \over {18}} = {{ - 4} \over 9}  \cr  & b)D = {{{2^{15}}{{.9}^3}} \over {{6^7}{{.4}^4}}} = {{{2^{15}}.{{({3^2})}^3}} \over {{{(2.3)}^7}.{{({2^2})}^4}}} = {{{2^{15}}{{.3}^6}} \over {{2^7}{{.3}^7}{{.2}^8}}} = {{{2^{15}}{{.3}^6}} \over {{2^{15}}{{.3}^7}}} = {1 \over 3}  \cr  & c)E = \left[ {{{\left( { - {1 \over 3}} \right)}^2}.{{27} \over 7} + \sqrt {{4 \over {49}}}  - 3} \right]:{4 \over 7} = \left( {{1 \over 9}.{{27} \over 7} + {2 \over 7} - 3} \right).{7 \over 4}  \cr  &  = \left( {{3 \over 7} + {2 \over 7} - {{21} \over 7}} \right).{7 \over 4} = {{ - 16} \over 7}.{7 \over 4} =  - 4 \cr} \)

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