Đáp án:
$\begin{array}{l}
B2)\\
a)\dfrac{{ - 12}}{{13}} = \dfrac{{ - 12.6}}{{78}} = \dfrac{{ - 72}}{{78}}\\
- \dfrac{1}{2} = \dfrac{{ - 39}}{{78}}\\
1 = \dfrac{{78}}{{78}}\\
\dfrac{{ - 4}}{3} = \dfrac{{ - 4.26}}{{78}} = \dfrac{{ - 104}}{{78}}\\
\dfrac{5}{3} = \dfrac{{5.26}}{{78}} = \dfrac{{130}}{{78}}\\
Do:\dfrac{{130}}{{78}} > \dfrac{{78}}{{78}} > \dfrac{{ - 39}}{{78}} > \dfrac{{ - 72}}{{78}} > \dfrac{{ - 104}}{{78}}\\
\Rightarrow \dfrac{5}{3} > 1 > \dfrac{{ - 1}}{2} > - \dfrac{{12}}{{13}} > - \dfrac{4}{3}\\
b)1 > \dfrac{3}{5} > - \dfrac{4}{7} > - \dfrac{5}{7}\\
\dfrac{{11}}{{12}} > \dfrac{3}{5}\\
\dfrac{{12}}{{13}} > \dfrac{{11}}{{12}}\\
\Rightarrow \dfrac{{12}}{{13}} > \dfrac{{11}}{{12}} > \dfrac{3}{5} > - \dfrac{4}{7} > - \dfrac{5}{7}\\
B3)\\
a)\dfrac{x}{3} + \dfrac{1}{5} = \dfrac{{2x}}{5} + \dfrac{2}{3}\\
\Rightarrow \dfrac{x}{3} - \dfrac{{2x}}{5} = \dfrac{2}{3} - \dfrac{1}{5}\\
\Rightarrow x.\left( {\dfrac{1}{3} - \dfrac{2}{5}} \right) = \dfrac{7}{{15}}\\
\Rightarrow x.\dfrac{{ - 1}}{{15}} = \dfrac{7}{{15}}\\
\Rightarrow x = - 7\\
b)\dfrac{1}{2} + \dfrac{2}{3} + \dfrac{3}{2} + ... + x = 2021.x\\
\Rightarrow \dfrac{1}{2}.\left( {1 + 2 + 3 + ... + 2x} \right) = 2021.x\\
\Rightarrow 1 + 2 + 3 + ... + 2x = 4042.x\\
\Rightarrow \dfrac{{\left( {2x + 1} \right).2x}}{2} = 4042.x\\
\Rightarrow 2x + 1 = 4042\\
\Rightarrow 2x = 4041\\
\Rightarrow x = \dfrac{{4041}}{2}\\
c)\dfrac{2}{{1.3}} + \dfrac{2}{{3.5}} + \dfrac{2}{{5.7}} + ... + \dfrac{2}{{\left( {2x + 1} \right)\left( {2x + 3} \right)}} = \dfrac{{102}}{{103}}\\
\Rightarrow 1 - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{5} + ... + \dfrac{1}{{2x + 1}} - \dfrac{1}{{2x + 3}} = \dfrac{{102}}{{103}}\\
\Rightarrow 1 - \dfrac{1}{{2x + 3}} = \dfrac{{102}}{{103}}\\
\Rightarrow \dfrac{1}{{2x + 3}} = \dfrac{1}{{103}}\\
\Rightarrow 2x + 3 = 103\\
\Rightarrow 2x = 100\\
\Rightarrow x = 50\\
d)x - 2x + 3x - 4x + ... + 9x - 10x = 1\\
\Rightarrow \left( { - x} \right) + \left( { - x} \right) + ... + \left( { - x} \right) = 1\\
\Rightarrow - 5x = 1\\
\Rightarrow x = - \dfrac{1}{5}
\end{array}$
