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