Cộng trừ các số hữu tỉ

Ta có:

$\begin{array}{l}\dfrac{1}{{1.4}} + \dfrac{1}{{4.7}} + \dfrac{1}{{7.10}} + ... + \dfrac{1}{{2020.2023}}\\ = \dfrac{1}{3}.\left( {\dfrac{3}{{1.4}} + \dfrac{3}{{4.7}} + \dfrac{3}{{7.10}} + ... + \dfrac{3}{{2020.2023}}} \right)\\ = \dfrac{1}{3}.\left( {\dfrac{1}{1} - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{7} + \dfrac{1}{7} - \dfrac{1}{{10}} + ... + \dfrac{1}{{2020}} - \dfrac{1}{{2023}}} \right)\\ = \dfrac{1}{3}.\left( {1 - \dfrac{1}{{2023}}} \right) = \dfrac{1}{3}.\dfrac{{2022}}{{2023}} = \dfrac{{674}}{{2023}}\end{array}$.

$\begin{array}{l}\left( {2022 + \dfrac{1}{{2022}} - 2021 - \dfrac{1}{{2021}}} \right).x = \dfrac{1}{4} + \dfrac{1}{{12}} - \dfrac{1}{3}\\\left( {1 + \dfrac{1}{{2022}} - \dfrac{1}{{2021}}} \right).x = \dfrac{3}{{12}} + \dfrac{1}{{12}} - \dfrac{4}{{12}}\\\left( {1 + \dfrac{1}{{2022}} - \dfrac{1}{{2021}}} \right).x = 0\end{array}$

Mà $1 + \dfrac{1}{{2022}} - \dfrac{1}{{2021}} > 0$ nên $x = 0$.

$\begin{array}{l}\dfrac{3}{{16}} - \left( {x + \dfrac{5}{8}} \right) = \dfrac{1}{{16}} - 1\\\dfrac{3}{{16}} - \left( {x + \dfrac{5}{8}} \right) = \dfrac{{ - 15}}{{16}}\\x + \dfrac{5}{8} = \dfrac{3}{{16}} - \left( {\dfrac{{ - 15}}{{16}}} \right)\\x + \dfrac{5}{8} = \dfrac{{18}}{{16}}\\x = \dfrac{{18}}{{16}} - \dfrac{5}{8}\\x = \dfrac{8}{{16}}\\x = \dfrac{1}{2}\end{array}$

Vậy $x = \dfrac{1}{2}$.

$\begin{array}{l}x - \left( {\dfrac{{ - 1}}{2}} \right) = 1\dfrac{1}{3} - \dfrac{1}{9}\\x + \dfrac{1}{2} = \dfrac{4}{3} - \dfrac{1}{9}\\x + \dfrac{1}{2} = \dfrac{{11}}{9}\\x = \dfrac{{11}}{9} - \dfrac{1}{2}\\x = \dfrac{{13}}{{18}}\end{array}$

Vậy $x = \dfrac{{13}}{{18}}$.

$\begin{array}{l}A = \left( {\dfrac{{12}}{7} - \dfrac{1}{2} + \dfrac{3}{5}} \right) - \left( {\dfrac{4}{5} + \dfrac{4}{7} - \dfrac{3}{2}} \right) - \left( {1 - \dfrac{1}{5} + \dfrac{1}{7}} \right)\\A = \dfrac{{12}}{7} - \dfrac{1}{2} + \dfrac{3}{5} - \dfrac{4}{5} - \dfrac{4}{7} + \dfrac{3}{2} - 1 + \dfrac{1}{5} - \dfrac{1}{7}\\A = \left( {\dfrac{{12}}{7} - \dfrac{4}{7} - \dfrac{1}{7}} \right) + \left( { - \dfrac{1}{2} + \dfrac{3}{2}} \right) + \left( {\dfrac{3}{5} - \dfrac{4}{5} + \dfrac{1}{5}} \right) - 1\\A = 1 + 1 + 0 - 1\\A = 1\end{array}$

$\begin{array}{l}\left( {2021 - \dfrac{1}{{2020}} - \dfrac{1}{3}} \right) - \left( {1 - \dfrac{1}{{2020}} - \dfrac{2}{3}} \right) - 1\dfrac{1}{3}\\ = 2021 - \dfrac{1}{{2020}} - \dfrac{1}{3} - 1 + \dfrac{1}{{2020}} + \dfrac{2}{3} - \dfrac{4}{3}\\ = (2021 - 1) + \left( { - \dfrac{1}{{2020}} + \dfrac{1}{{2020}}} \right) + \left( { - \dfrac{1}{3} + \dfrac{2}{3} - \dfrac{4}{3}} \right)\\ = 2020 + 0 + ( - 1)\\ = 2019\end{array}$

$D = \dfrac{{2019}}{{2022}} - \dfrac{2}{{2021}} + \dfrac{3}{{2022}} - \dfrac{{4040}}{{2021}}$

$D = \left( {\dfrac{{2019}}{{2022}} + \dfrac{3}{{2022}}} \right) - \left( {\dfrac{2}{{2021}} + \dfrac{{4040}}{{2021}}} \right)$

$D = \dfrac{{2022}}{{2022}} - \dfrac{{4042}}{{2021}}$

$D = 1 - 2= -1$.

$\dfrac{1}{2} + \dfrac{1}{6} = \dfrac{{1.3}}{{2.3}} + \dfrac{1}{6} = \dfrac{3}{6} + \dfrac{1}{6} = \dfrac{{3 + 1}}{6} = \dfrac{4}{6} = \dfrac{2}{3}$.

Ta có: $\dfrac{5}{{12}} + \dfrac{{ - 7}}{{24}} = \dfrac{{5.2}}{{12.2}} + \dfrac{{ - 7}}{{24}} = \dfrac{{10}}{{24}} + \dfrac{{ - 7}}{{24}} = \dfrac{{10 + ( - 7)}}{{24}} = \dfrac{3}{{24}} = \dfrac{1}{8} > 0$.

Do đó kết quả là số hữu tỉ dương.

Ta có: $\dfrac{{11}}{5} + \dfrac{2}{7} = \dfrac{{77}}{{35}} + \dfrac{{10}}{{35}} = \dfrac{{77 + 10}}{{35}} = \dfrac{{87}}{{35}}$;

$\dfrac{{11}}{7} + \dfrac{2}{5} = \dfrac{{55}}{{35}} + \dfrac{{14}}{{35}} = \dfrac{{55 + 14}}{{35}} = \dfrac{{69}}{{35}}$;

$\dfrac{3}{7} + \dfrac{9}{5} = \dfrac{{15}}{{35}} + \dfrac{{63}}{{35}} = \dfrac{{15 + 63}}{{35}} = \dfrac{{78}}{{35}}$;

$1 + \dfrac{{33}}{{35}} = \dfrac{{35}}{{35}} + \dfrac{{33}}{{35}} = \dfrac{{35 + 33}}{{35}} = \dfrac{{68}}{{35}}$.

Do đó $\dfrac{{69}}{{35}}$ là kết quả của phép tính: $\dfrac{{11}}{7} + \dfrac{2}{5}$.

Ta có: $\dfrac{7}{3} - \dfrac{{23}}{5} = \dfrac{{35}}{{15}} - \dfrac{{69}}{{15}} = \dfrac{{35 - 69}}{{15}} = \dfrac{{ - 34}}{{15}}$;

$\dfrac{5}{3} - \dfrac{3}{5} = \dfrac{{25}}{{15}} - \dfrac{9}{{15}} = \dfrac{{25 - 9}}{{15}} = \dfrac{{16}}{{15}}$;

$\dfrac{3}{5} - \dfrac{5}{3} = \dfrac{9}{{15}} - \dfrac{{25}}{{15}} = \dfrac{{9 - 25}}{{15}} = \dfrac{{ - 16}}{{15}}$;

$\dfrac{{18}}{5} - \dfrac{2}{3} = \dfrac{{54}}{{15}} - \dfrac{{10}}{{15}} = \dfrac{{54 - 10}}{{15}} = \dfrac{{44}}{{15}}$.   

Vậy số $\dfrac{{16}}{{15}}$ viết thành hiệu của hai số hữu tỉ dương  $\dfrac{5}{3} - \dfrac{3}{5}$.

$\begin{array}{l}M = 2021 - \left[ {\left( {\dfrac{1}{3} + \dfrac{1}{{12}} - \dfrac{5}{6}} \right) + 2020} \right]\\M = 2021 - \left[ {\left( {\dfrac{4}{12} + \dfrac{1}{{12}} - \dfrac{{10}}{{12}}} \right) + 2020} \right]\\M = 2021 - \left( { - \dfrac{5}{{12}} + 2020} \right)\\M = 2021 + \dfrac{5}{{12}} - 2020\\M = \dfrac{5}{{12}}+1=\dfrac{17}{{12}}\end{array}$

Ta thấy: $M = \dfrac{17}{{12}} > \dfrac{{12}}{{12}} = 1$ nên $M > 1$

$\dfrac{5}{{11}} + \dfrac{9}{{20}} + \left( { - \dfrac{5}{{11}}} \right) = \left[ {\dfrac{5}{{11}} + \left( { - \dfrac{5}{{11}}} \right)} \right] + \dfrac{9}{{20}} = \dfrac{{5 + ( - 5)}}{{11}} + \dfrac{9}{{20}} = 0 + \dfrac{9}{{20}} = \dfrac{9}{{20}}$.

$\begin{array}{l}N = \dfrac{{ - 2}}{3} - \dfrac{4}{5} - \left( { - \dfrac{3}{{20}}} \right)\\N = \dfrac{{ - 2}}{3} - \dfrac{4}{5} + \dfrac{3}{{20}}\\N = \dfrac{{ - 40}}{{60}} - \dfrac{{48}}{{60}} + \dfrac{9}{{60}}\\N = \dfrac{{ - 79}}{{60}}\end{array}$

$x + \dfrac{3}{7} =  - \dfrac{3}{{14}}$

$x =  - \dfrac{3}{{14}} - \dfrac{3}{7}$

$x = \dfrac{{ - 3}}{{14}} - \dfrac{6}{{14}}$

$x = \dfrac{{ - 3 - 6}}{{14}}$

$x = \dfrac{{ - 9}}{{14}}$

$\begin{array}{l}x - \dfrac{1}{3} = \frac{2}{9}\\x = \dfrac{2}{9} + \dfrac{1}{3}\\x = \dfrac{2}{9} + \dfrac{3}{9}\\x = \dfrac{5}{9}\end{array}$

Vậy $x = \dfrac{5}{9}$.

Ta có $\dfrac{4}{5} - \left( { - \dfrac{2}{7}} \right) + \left( { - \dfrac{5}{{10}}} \right)$$= \dfrac{4}{5} + \dfrac{2}{7} + \dfrac{{ - 5}}{{10}}$$= \dfrac{{4.14}}{{5.14}} + \dfrac{{2.10}}{{7.10}} + \dfrac{{ - 5.7}}{{10.7}} = \dfrac{{56}}{{70}} + \dfrac{{20}}{{70}} + \dfrac{{ - 35}}{{70}} = \dfrac{{56 + 20 + ( - 35)}}{{70}} = \dfrac{{41}}{{70}}$.

Ta có: $B = \left( {\dfrac{1}{2} - \dfrac{7}{{13}} - \dfrac{1}{3}} \right) + \left( { - \dfrac{6}{{13}} + \dfrac{1}{2} + 1\dfrac{1}{3}} \right)$

$= \dfrac{1}{2} - \dfrac{7}{{13}} - \dfrac{1}{3} - \dfrac{6}{{13}} + \dfrac{1}{2} + \dfrac{4}{3}$

$\begin{array}{l} = \dfrac{1}{2} + \dfrac{{ - 7}}{{13}} + \dfrac{{ - 1}}{3} + \dfrac{{ - 6}}{{13}} + \dfrac{1}{2} + \dfrac{4}{3}\\ = \left( {\dfrac{1}{2} + \dfrac{1}{2}} \right) + \left( {\dfrac{{ - 7}}{{13}} + \dfrac{{ - 6}}{{13}}} \right) + \left( {\dfrac{{ - 1}}{3} + \dfrac{4}{3}} \right)\end{array}$

$= \dfrac{{1 + 1}}{2} + \dfrac{{( - 7) + ( - 6)}}{{13}} + \dfrac{{ - 1 + 4}}{3}$

$= \dfrac{2}{2} + \dfrac{{ - 13}}{{13}} + \dfrac{3}{3}$

$= 1 + ( - 1) + 1$

$= 1$

Do $0 < 1 < 2$ nên $0 < B < 2$.

$\begin{array}{l}\dfrac{1}{{5}} - x = \dfrac{7}{{15}}\\x = \dfrac{1}{{5}} - \dfrac{7}{{15}}\\x = \dfrac{{ - 4}}{{15}}\end{array}$

Vậy $x = \dfrac{{ - 4}}{{15}}$.