Đáp án:
$\begin{array}{l}
d)\dfrac{{13}}{4} - \dfrac{{11}}{5}:\dfrac{2}{{10}} + \dfrac{3}{4} \le x \le \dfrac{5}{{14}}:\dfrac{{15}}{{14}} + 1\left( {x \in Z} \right)\\
\Leftrightarrow \dfrac{{13}}{4} - 11 + \dfrac{3}{4} \le x \le \dfrac{1}{3} + 1\\
\Leftrightarrow - 7 \le x \le \dfrac{4}{3}\\
\Leftrightarrow x \in \left\{ { - 7; - 6; - 5;...;1} \right\}\\
Vậy\,x \in \left\{ { - 7; - 6; - 5;...;1} \right\}\\
B2)\\
A = \dfrac{2}{{3.5}} + \dfrac{2}{{5.7}} + \dfrac{2}{{7.9}} + ... + \dfrac{2}{{37.39}}\\
= \dfrac{1}{3} - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{7} + \dfrac{1}{7} - \dfrac{1}{9} + ... + \dfrac{1}{{37}} - \dfrac{1}{{39}}\\
= \dfrac{1}{3} - \dfrac{1}{{39}}\\
= \dfrac{{12}}{{39}}\\
B = \dfrac{3}{{2.5}} + \dfrac{3}{{5.8}} + \dfrac{3}{{8.11}} + ... + \dfrac{3}{{32.35}}\\
= \dfrac{1}{2} - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{8} + \dfrac{1}{8} - \dfrac{1}{{11}} + ... + \dfrac{1}{{32}} - \dfrac{1}{{35}}\\
= \dfrac{1}{2} - \dfrac{1}{{35}}\\
= \dfrac{{33}}{{70}}
\end{array}$
