Đáp án:
$\begin{array}{l}
a)A = \dfrac{7}{{35}}.\dfrac{{10}}{9} + \dfrac{7}{{19}}.\dfrac{9}{{35}} - \dfrac{2}{{35}}\\
= \dfrac{7}{{35}}.\dfrac{{10}}{9} + \dfrac{7}{{35}}.\dfrac{9}{{35}} - \dfrac{7}{{35}}.\dfrac{2}{7}\\
= \dfrac{7}{{35}}\left( {\dfrac{{10}}{9} + \dfrac{9}{{19}} - \dfrac{2}{7}} \right)\\
= \dfrac{7}{{35}}.\dfrac{{1555}}{{1197}}\\
= \dfrac{{311}}{{1197}}\\
b)B = \dfrac{{{{5.4}^{15}}{{.9}^9} - {{4.3}^{20}}{{.8}^9}}}{{{{5.2}^{10}}{{.6}^{19}} - {{7.2}^{29}}{{.27}^6}}}\\
= \dfrac{{{{5.2}^{30}}{{.3}^{18}} - {2^{29}}{{.3}^{20}}}}{{{{5.2}^{29}}{{.3}^{19}} - {{7.2}^{29}}{{.3}^{18}}}}\\
= \dfrac{{{2^{29}}{{.3}^{18}}(5.2 - 9)}}{{{2^{29}}{{.3}^{18}}.(5.3 - 7.2)}}\\
= 1\\
c)C = \left( {1 + \dfrac{1}{{1.3}}} \right)\left( {1 + \dfrac{1}{{2.4}}} \right)\left( {1 + \dfrac{1}{{3.5}}} \right)...\left( {1 + \dfrac{1}{{98.100}}} \right)\\
= \dfrac{{{2^2}}}{{1.3}}.\dfrac{{{3^{^2}}}}{{2.4}}.\dfrac{{{4^2}}}{{3.5}}...\dfrac{{{{99}^2}}}{{98.100}}\\
= \dfrac{{2.3.4...99}}{{1.2.3...98}}.\dfrac{{2.3.4...99}}{{3.4.5...100}}\\
= 99.\dfrac{2}{{100}}\\
= \dfrac{{198}}{{100}}\\
d)D = \dfrac{{155 - \dfrac{{10}}{7} - \dfrac{5}{{11}} + \dfrac{5}{{23}}}}{{403 - \dfrac{{26}}{7} - \dfrac{{13}}{5} + \dfrac{{13}}{{23}}}} + \dfrac{{\dfrac{3}{5} + \dfrac{3}{{13}} - 0,9}}{{\dfrac{7}{{91}} + 0,2 - \dfrac{3}{{10}}}}\\
= \dfrac{{5\left( {31 - \dfrac{2}{7} - \dfrac{1}{{11}} + \dfrac{1}{{23}}} \right)}}{{13\left( {31 - \dfrac{2}{7} - \dfrac{1}{{11}} + \dfrac{1}{{23}}} \right)}} + \dfrac{{\dfrac{3}{5} + \dfrac{3}{{13}} - \dfrac{9}{{10}}}}{{\dfrac{1}{5} + \dfrac{1}{{13}} - \dfrac{3}{{10}}}}\\
= \dfrac{5}{{13}} + 3\\
= \dfrac{{44}}{{13}}
\end{array}$