Đáp án:
$\begin{array}{l}
B2)\\
a)\dfrac{{17}}{6} - \left( {x - \dfrac{7}{6}} \right) = \dfrac{7}{4}\\
\Leftrightarrow \dfrac{{17}}{6} - x + \dfrac{7}{6} = \dfrac{7}{4}\\
\Leftrightarrow x = \dfrac{{24}}{6} - \dfrac{7}{4}\\
\Leftrightarrow x = 4 - \dfrac{7}{4}\\
\Leftrightarrow x = \dfrac{9}{4}\\
Vậy\,x = \dfrac{9}{4}\\
b)\dfrac{4}{3} + \left( {1,25 - x} \right) = 2,25\\
\Leftrightarrow x = \dfrac{4}{3} + 1,25 - 2,25\\
\Leftrightarrow x = \dfrac{4}{3} - 1\\
\Leftrightarrow x = \dfrac{1}{3}\\
Vậy\,x = \dfrac{1}{3}\\
c)2x - 3 = x + \dfrac{1}{2}\\
\Leftrightarrow 2x - x = \dfrac{1}{2} + 3\\
\Leftrightarrow x = \dfrac{7}{2}\\
Vậy\,x = \dfrac{7}{2}\\
d)4x - \left( {2x + 1} \right) = 3 - \dfrac{1}{3} + x\\
\Leftrightarrow 4x - 2x - x = 3 - \dfrac{1}{3} + 1\\
\Leftrightarrow x = \dfrac{{11}}{3}\\
Vậy\,x = \dfrac{{11}}{3}\\
B3)\\
a)\dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + ... + \dfrac{1}{{2019.2020}}\\
= 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{{2019}} - \dfrac{1}{{2020}}\\
= 1 - \dfrac{1}{{2020}}\\
= \dfrac{{2019}}{{2020}}\\
b)\dfrac{1}{{1.4}} + \dfrac{1}{{4.7}} + \dfrac{1}{{7.10}} + ... + \dfrac{1}{{100.103}}\\
= \dfrac{1}{3}.\left( {\dfrac{3}{{1.4}} + \dfrac{3}{{4.7}} + \dfrac{3}{{7.10}} + ... + \dfrac{3}{{100.103}}} \right)\\
= \dfrac{1}{3}.\left( {1 - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{7} + \dfrac{1}{7} - \dfrac{1}{{10}} + ... + \dfrac{1}{{100}} - \dfrac{1}{{103}}} \right)\\
= \dfrac{1}{3}.\left( {1 - \dfrac{1}{{103}}} \right)\\
= \dfrac{1}{3}.\dfrac{{102}}{{103}}\\
= \dfrac{{34}}{{103}}
\end{array}$
