Đề bài
Tính:
\[\eqalign{ & a]\,\,C = {{{8^{10}} + {4^{10}}} \over {{8^4} + {4^{11}}}} \cr & b]\,\,D = {1 \over {2.3}} + {1 \over {3.4}} + {1 \over {4.5}} + ...... + {1 \over {2015.2016}} \cr & c]\,\,E = {1 \over 3} - {3 \over 4} + {3 \over 5} + {2 \over {2015}} - {1 \over {36}} + {1 \over {15}} - {2 \over 9} \cr} \]
Lời giải chi tiết
\[\eqalign{ & a]C = {{{8^{10}} + {4^{10}}} \over {{8^4} + {4^{11}}}} = {{{{[{2^3}]}^{10}} + {{[{2^2}]}^{10}}} \over {{{[{2^3}]}^4} + {{[{2^2}]}^{11}}}} = {{{2^{20}}[{2^{10}} + 1]} \over {{2^{12}}[1 + {2^{10}}]}} = {{{2^{20}}} \over {{2^{12}}}} = {2^8} = 256 \cr & b]D = {1 \over {2.3}} + {1 \over {3.4}} + {1 \over {4.5}} + ... + {1 \over {2015.2016}} \cr & = {1 \over 2} - {1 \over 3} + {1 \over 3} - {1 \over 4} + {1 \over 4} - {1 \over 5} + ... + {1 \over {2015}} - {1 \over {2016}} = {1 \over 2} - {1 \over {2016}} = {{1007} \over {2016}} \cr & c]E = {1 \over 3} - {3 \over 4} + {3 \over 5} + {2 \over {2015}} - {1 \over {36}} + {1 \over {15}} - {2 \over 9} = {1 \over 3} + {3 \over 5} + {1 \over {15}} - {3 \over 4} - {1 \over {36}} - {2 \over 9} + {2 \over {2015}} \cr & = \left[ {{5 \over {15}} + {9 \over {15}} + {1 \over {15}}} \right] + \left[ {{{ - 27} \over {36}} - {1 \over {36}} - {8 \over {36}}} \right] + {2 \over {2015}} \cr & = {{15} \over {15}} + \left[ {{{ - 36} \over {36}}} \right] + {2 \over {2015}} = 1 + [ - 1] + {2 \over {2015}} = {2 \over {2015}} \cr} \]