Cộng trừ các số hữu tỉ

$B = \dfrac{{19}}{{ - 18}} + \dfrac{{11}}{{15}} + \dfrac{1}{{18}} + \dfrac{4}{{15}} + 4$

$= \dfrac{{ - 19}}{{18}} + \dfrac{{11}}{{15}} + \dfrac{1}{{18}} + \dfrac{4}{{15}} + 4$

$= \left( {\dfrac{{ - 19}}{{18}} + \dfrac{1}{{18}}} \right) + \left( {\dfrac{{11}}{{15}} + \dfrac{4}{{15}}} \right) + 4$

$= \dfrac{{ - 19 + 1}}{{18}} + \dfrac{{11 + 4}}{{15}} + 4$

$= \dfrac{{ - 18}}{{18}} + \dfrac{{15}}{{15}} + 4$

$=  - 1 + 1 + 4$

$= 4$

$\dfrac{9}{{2021}} - \dfrac{1}{3} + \left( {\dfrac{{ - 9}}{{2021}}} \right) = \left( {\dfrac{9}{{2021}} - \dfrac{9}{{2021}}} \right) - \dfrac{1}{3} = 0 - \dfrac{1}{3} = \dfrac{{ - 1}}{3}$.

$x + y = a + \dfrac{b}{{2c}} = \dfrac{{a.2c}}{{2c}} + \dfrac{b}{{2c}} = \dfrac{{2ac + b}}{{2c}}$.

$\left( {\dfrac{3}{8} - \dfrac{1}{5} - \dfrac{1}{3}} \right) + \left( {\dfrac{5}{8} - 1\dfrac{3}{7} - \dfrac{4}{7}} \right) + \dfrac{1}{3}$ $= \left( {\dfrac{3}{8} - \dfrac{1}{5} - \dfrac{1}{3}} \right) + \left( {\dfrac{5}{8} - \dfrac{{10}}{7} - \dfrac{4}{7}} \right) + \dfrac{1}{3}$ $= \dfrac{3}{8} - \dfrac{1}{5} - \dfrac{1}{3} + \dfrac{5}{8} - \dfrac{{10}}{7} - \dfrac{4}{7} + \dfrac{1}{3}$

$= \dfrac{3}{8} + \dfrac{{ - 1}}{5} + \dfrac{{ - 1}}{3} + \dfrac{5}{8} + \dfrac{{ - 10}}{7} + \dfrac{{ - 4}}{7} + \dfrac{1}{3}$ $= \left( {\dfrac{3}{8} + \dfrac{5}{8}} \right) + \left( {\dfrac{{ - 1}}{3} + \dfrac{1}{3}} \right) + \left( {\dfrac{{ - 10}}{7} + \dfrac{{ - 4}}{7}} \right) + \dfrac{{ - 1}}{5}$

$= \dfrac{8}{8} + \dfrac{0}{3} + \dfrac{{ - 14}}{7} + \dfrac{{ - 1}}{5}$ $= 1 + 0 + ( - 2) + \dfrac{{ - 1}}{5}$ $=  - 1 + \dfrac{{ - 1}}{5} = \dfrac{{ - 5}}{5} + \dfrac{{ - 1}}{5} = \dfrac{{ - 6}}{5}$.

Ta có: $0,75 = \dfrac{{75}}{{100}} = \dfrac{3}{4};\,1\dfrac{2}{5} = \dfrac{7}{5}$.

Do đó

$D = \dfrac{3}{4} + \dfrac{2}{5} + \left( {\dfrac{1}{9} - \dfrac{7}{5} + \dfrac{5}{4}} \right) - \left( {\dfrac{1}{9} + 1} \right)$

$= \dfrac{3}{4} + \dfrac{2}{5} + \dfrac{1}{9} - \dfrac{7}{5} + \dfrac{5}{4} - \dfrac{1}{9} - 1$

$\begin{array}{l} = \dfrac{3}{4} + \dfrac{2}{5} + \dfrac{1}{9} + \dfrac{{ - 7}}{5} + \dfrac{5}{4} + \dfrac{{ - 1}}{9} + ( - 1)\\ = \left( {\dfrac{3}{4} + \dfrac{5}{4}} \right) + \left( {\dfrac{2}{5} + \dfrac{{ - 7}}{5}} \right) + \left( {\dfrac{1}{9} + \dfrac{{ - 1}}{9}} \right) + ( - 1)\end{array}$

$\begin{array}{l} = \dfrac{8}{4} + \dfrac{{ - 5}}{5} + 0 + ( - 1)\\ = 2 + ( - 1) + ( - 1)\\ = 0\end{array}$

Vậy $D = 0$.

Ta có: $- x + \dfrac{1}{7} = 3 - 1\dfrac{1}{2}$

$- x + \dfrac{1}{7} = 3 - \dfrac{3}{2}$

$- x + \dfrac{1}{7} = \dfrac{6}{2} - \dfrac{3}{2}$

$- x + \dfrac{1}{7} = \dfrac{3}{2}$

$x = \dfrac{1}{7} - \dfrac{3}{2}$

$x = \dfrac{2}{{14}} - \dfrac{{21}}{{14}}$

$x = \dfrac{{ - 19}}{{14}}$

Vậy $x = \dfrac{{ - 19}}{{14}}$.

Ta có:

$1 - \left( {\dfrac{8}{7} - x} \right) = \dfrac{4}{5} + \left( { - \dfrac{1}{{10}}} \right)$

$1 - \left( {\dfrac{8}{7} - x} \right) = \dfrac{8}{{10}} + \dfrac{{ - 1}}{{10}}$

$1 - \left( {\dfrac{8}{7} - x} \right) = \dfrac{7}{{10}}$

$\dfrac{8}{7} - x = 1 - \dfrac{7}{{10}}$

$\dfrac{8}{7} - x = \dfrac{{10}}{{10}} - \dfrac{7}{{10}}$

$\dfrac{8}{7} - x = \dfrac{3}{{10}}$

$x = \dfrac{8}{7} - \dfrac{3}{{10}}$

$x = \dfrac{{80}}{{70}} - \dfrac{{21}}{{70}}$

$x = \dfrac{{59}}{{70}}$

Vậy $x = \dfrac{{59}}{{70}}$.

Quan sát đáp án ta thấy A và D là hiệu của một số hữu tỉ âm và một số hữu tỉ dương nên không thỏa mãn.

Đáp án B có: $\dfrac{3}{7} - \dfrac{2}{3} = \dfrac{{ - 5}}{{21}}$ => B thỏa mãn

Đáp án C có: $\dfrac{5}{7} - \dfrac{4}{3} = \dfrac{{ - 13}}{{21}} \ne \dfrac{{ - 5}}{{21}}$=> C không thỏa mãn.

$\left( {2020 + \dfrac{1}{{2020}} - 2021 - \dfrac{1}{{2021}}} \right)\left( {x - \dfrac{1}{2}} \right) = \dfrac{1}{3} - \dfrac{2}{{15}} - \dfrac{1}{5}$

$\left( {2020 + \dfrac{1}{{2020}} - 2021 - \dfrac{1}{{2021}}} \right)\left( {x - \dfrac{1}{2}} \right) = \dfrac{5}{{15}} - \dfrac{2}{{15}} - \dfrac{3}{{15}}$

$\left( {2020 + \dfrac{1}{{2020}} - 2021 - \dfrac{1}{{2021}}} \right)\left( {x - \dfrac{1}{2}} \right) = 0$

Mà $2020 + \dfrac{1}{{2020}} - 2021 - \dfrac{1}{{2021}} = \left( {\dfrac{1}{{2020}} - \dfrac{1}{{2021}}} \right) + (2020 - 2021) = \dfrac{1}{{2020.2021}} - 1 < 0$  (do $\dfrac{1}{{2020.2021}} < 1$)

Do đó $x - \dfrac{1}{2} = 0$ hay $x = \dfrac{1}{2}$

Vậy ${x_0} = \dfrac{1}{2}$.

$\dfrac{1}{{1.3}} + \dfrac{1}{{3.5}} + \dfrac{1}{{5.7}} + \dfrac{1}{{7.9}} + ... + \dfrac{1}{{2021.2023}}$

$= \dfrac{1}{2}.\left( {\dfrac{2}{{1.3}} + \dfrac{2}{{3.5}} + \dfrac{2}{{5.7}} + \dfrac{2}{{7.9}} + ... + \dfrac{2}{{2021.2023}}} \right)$

$= \dfrac{1}{2}.\left( {1 - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{7} + \dfrac{1}{7} - \dfrac{1}{9} + ... - \dfrac{1}{{2021}} + \dfrac{1}{{2021}} - \dfrac{1}{{2023}}} \right)$

$\begin{array}{l} = \dfrac{1}{2}.\left( {1 - \dfrac{1}{{2023}}} \right)\\ = \dfrac{1}{2}.\left( {\dfrac{{2023}}{{2023}} - \dfrac{1}{{2023}}} \right)\\ = \dfrac{1}{2}.\dfrac{{2022}}{{2023}}\\ = \dfrac{{1011}}{{2023}}\end{array}$.

$1 + \dfrac{4}{5} = \dfrac{5}{5} + \dfrac{4}{5} = \dfrac{9}{5}$

$\dfrac{1}{3} - \dfrac{3}{7} = \dfrac{7}{{21}} - \dfrac{9}{{21}} = \dfrac{{ - 2}}{{21}}$.

Ta thấy $\dfrac{{ - 2}}{{21}}$ là một số hữu tỉ âm.

Ta có:

$1 - \dfrac{1}{{11}} = \dfrac{{11}}{{11}} - \dfrac{1}{{11}} = \dfrac{{10}}{{11}}$

$1 - \dfrac{{10}}{{11}} = \dfrac{{11}}{{11}} - \dfrac{{10}}{{11}} = \dfrac{1}{{11}}$

$1 - \dfrac{9}{{11}} = \dfrac{{11}}{{11}} - \dfrac{9}{{11}} = \dfrac{2}{{11}}$

$1 - \dfrac{2}{{11}} = \dfrac{{11}}{{11}} - \dfrac{2}{{11}} = \dfrac{9}{{11}}$

Vậy $\dfrac{1}{{11}}$ là kết quả của $1 - \dfrac{{10}}{{11}}$.

Ta có:

$\dfrac{1}{3} - \dfrac{1}{{12}} = \dfrac{4}{{12}} - \dfrac{1}{{12}} = \dfrac{3}{{12}} = \dfrac{1}{4}$

$\dfrac{1}{3} + \dfrac{1}{{12}} = \dfrac{4}{{12}} + \dfrac{1}{{12}} = \dfrac{5}{{12}}$

$\dfrac{1}{3} - \dfrac{1}{4} = \dfrac{4}{{12}} - \dfrac{3}{{12}} = \dfrac{1}{{12}}$

$\dfrac{1}{3} + \dfrac{1}{4} = \dfrac{4}{{12}} + \dfrac{3}{{12}} = \dfrac{7}{{12}}$

Vậy $\dfrac{7}{{12}}$ là kết quả của $\dfrac{1}{3} + \dfrac{1}{{4}}$.

 $\dfrac{3}{4} - 1 = \dfrac{3}{4} - \dfrac{4}{4} = \dfrac{{ - 1}}{4}$

$\dfrac{3}{4} + 1 = \dfrac{3}{4} + \dfrac{4}{4} = \dfrac{7}{4}$

$\dfrac{{ - 3}}{4} + 1 = \dfrac{{ - 3}}{4} + \dfrac{4}{4} = \dfrac{1}{4}$

$\dfrac{{ - 3}}{4} - 1 = \dfrac{{ - 3}}{4} - \dfrac{4}{4} = \dfrac{{ - 7}}{4}$

Vậy $\dfrac{3}{4} - 1$ có kết quả là $\dfrac{{ - 1}}{4}$.

Ta có: $\dfrac{4}{{12}} + \dfrac{{ - 3}}{{18}} = \dfrac{2}{6} + \dfrac{{ - 1}}{6} = \dfrac{1}{6}$.

$\dfrac{{ - 3}}{7} + \dfrac{{10}}{7} = \dfrac{{ - 3 + 10}}{7} = \dfrac{7}{7} = 1$.

$\dfrac{5}{9} + \left( {\dfrac{{ - 2}}{9}} \right) = \dfrac{{5 + ( - 2)}}{9} = \dfrac{{5 - 2}}{9} = \dfrac{3}{9}=\dfrac{1}{3}$.

$\dfrac{5}{6} + \left( {\dfrac{{ -11}}{6}} \right) = \dfrac{{5 + ( -11)}}{6} = \dfrac{{5 +(-11)}}{6} = \dfrac{-6}{6}=-1$