\( \displaystyle A = 11{3 \over {13}} - \left( {2{4 \over 7} + 5{3 \over {13}}} \right) \)\( \displaystyle = \left( {11{3 \over {13}} - 5{3 \over {13}}} \right) - 2{4 \over 7}\)
\( \displaystyle = 6 - 2{4 \over 7} = 6 - {{18} \over 7}\)\( \displaystyle = {{24} \over 7} = 3{3 \over 7}\)
\( \displaystyle B = \left( {6{4 \over 9} + 3{7 \over {11}}} \right) - 4{4 \over 9} \)\( \displaystyle = \left( {6{4 \over 9} - 4{4 \over 9}} \right) + 3{7 \over {11}}\)
\( \displaystyle = 2 + {{40} \over {11}} = {{62} \over {11}} \)\( \displaystyle = 5{7 \over {11}}\)
\( \displaystyle C = {{ - 5} \over 7}.{2 \over {11}} + {{ - 5} \over 7}.{9 \over {11}} + 1{5 \over 7} \)\( \displaystyle = {{ - 5} \over 7}\left( {{2 \over {11}} + {9 \over {11}}} \right) + 1{5 \over 7}\)
\( \displaystyle = {{ - 5} \over 7} + 1{5 \over 7} = {{ - 5} \over 7} + {{12} \over 7} \)\( \displaystyle = {7 \over 7} = 1\)
\( \displaystyle D = 0,7.2{2 \over 3}.20.0,375.{5 \over {28}} \)\( \displaystyle = {7 \over {10}}.{8 \over 3}.20.{{375} \over {1000}}.{5 \over {28}} \) \( = \dfrac{7}{{10}}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{{28}}\)
\( = \dfrac{7}{{10}}.\dfrac{5}{{28}}.\dfrac{8}{3}.\dfrac{3}{8}.20\)\( = \dfrac{1}{8}.1.20\)
\( \displaystyle = {{20} \over {8}} = {5 \over 2}\)
\( \displaystyle E = \left( { - 6,17 + 3{5 \over 9} - 2{{36} \over {97}}} \right).\left( {{1 \over 3} - 0,25 - {1 \over {12}}} \right)\)
Vì: \( \displaystyle {1 \over 3} - 0,25 - {1 \over {12}} = {1 \over 3} - {1 \over 4} - {1 \over {12}} \)\( \displaystyle = {{4 - 3 - 1} \over {12}} = 0\)
Trong tích E có một thừa số bằng 0 nên \(E = 0.\)
