II. PHẦN TỰ LUẬN

1. Thực hiện phép tính (tính nhanh nếu có thể):

a) \[\frac{3}{7} + \frac{{ - 1}}{5} + \frac{{ - 3}}{7}\];                                              b) \[\frac{7}{{19}}.\frac{8}{{11}} + \frac{7}{{19}}.\frac{3}{{11}} + \frac{{12}}{{19}}\];

2. Tìm \[x\]:

a) \[\frac{1}{2} - x = \frac{{ - 1}}{6}\];                                                 b) \[50\% \] của \[x\] bằng \[\frac{1}{5}\] của 15.

Ta có: \[A = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{{2022}}\];

\[B = \frac{{2021}}{1} + \frac{{2020}}{2} + \frac{{2019}}{3} + ... + \frac{1}{{2021}}\]

 \[ = 2021 + \frac{{2020}}{2} + \frac{{2019}}{3} + ... + \frac{1}{{2021}}\]

\[ = 2020 + \frac{{2020}}{2} + \frac{{2019}}{3} + ... + \frac{1}{{2021}} + 1\]

\[ = \,\left( {1 + \frac{{2020}}{2}} \right) + \left( {1 + \frac{{2019}}{3}} \right) + ... + \left( {\frac{1}{{2021}} + 1} \right) + 1\]

\[ = \,\frac{{2022}}{2} + \frac{{2022}}{3} + ... + \frac{{2022}}{{2021}} + \frac{{2022}}{{2022}}\]

\[ = 2022\,\,.\,\,\left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2021}} + \frac{1}{{2022}}} \right)\].

Khi đó,  \[\frac{B}{A} = \frac{{2022\,\,.\,\,\left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2021}} + \frac{1}{{2022}}} \right)}}{{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{{2022}}}} = 2022\].

Vậy tỉ số \[\frac{B}{A} = 2022\].

Đáp án đúng là: D

• \[\frac{3}{5} = \frac{{27}}{{45}}\] vì  \[3\,\,.\,\,45 = 5\,\,.\,\,27 = 135\];

• \[\frac{{ - 3}}{5} = \frac{3}{{ - 5}}\] vì \[\left( { - 3} \right)\,\,.\,\,\left( { - 5} \right) = 5\,\,.\,\,3 = 15\];

• \[\frac{{18}}{{27}} = \frac{{ - 2}}{{ - 3}}\] vì \[18\,\,.\,\,\left( { - 3} \right) = 27\,.\,\,\left( { - 2} \right) =  - \,54\];

• \[\frac{{17}}{{51}}\] không phải là phân số tối giản vì \[\frac{{17}}{{51}} = \frac{{17}}{{17\,.\,3}} = \frac{1}{3}\].