PHẦN II. TỰ LUẬN (7,0 điểm)

(1,5 điểm) Tính giá trị của các biểu thức sau (tính hợp lí nếu có thể):

a) \(\frac{4}{3} - \frac{1}{3}.\frac{2}{5}\);   

b) \(\sqrt {\frac{{196}}{{121}}} :\frac{7}{{11}} - \left| {\frac{{ - 5}}{4}} \right|\);                                                          

c) \(\left( {\frac{{ - 3}}{4} + \frac{4}{{15}}} \right):\frac{{2022}}{{2023}} + \left( {\frac{{ - 1}}{4} + \frac{{11}}{{15}}} \right):\frac{{2022}}{{2023}}\).

a) \(\frac{4}{3} - \frac{1}{3}.\frac{2}{5} = \frac{4}{3} - \frac{2}{{15}} = \frac{{20}}{{15}} - \frac{2}{{15}} = \frac{{18}}{{15}} = \frac{6}{5}\).

b) \(\sqrt {\frac{{196}}{{121}}} :\frac{7}{{11}} - \left| {\frac{{ - 5}}{4}} \right| = \frac{{14}}{{11}}.\frac{{11}}{7} - \frac{5}{4} = 2 - \frac{5}{4} = \frac{3}{4}\).

c) \(\left( {\frac{{ - 3}}{4} + \frac{4}{{15}}} \right):\frac{{2022}}{{2023}} + \left( {\frac{{ - 1}}{4} + \frac{{11}}{{15}}} \right):\frac{{2022}}{{2023}}\)

\( = \left( {\frac{{ - 3}}{4} + \frac{4}{{15}}} \right).\frac{{2023}}{{2022}} + \left( {\frac{{ - 1}}{4} + \frac{{11}}{{15}}} \right).\frac{{2023}}{{2022}}\)

\[ = \left( {\frac{{ - 3}}{4} + \frac{4}{{15}} + \frac{{ - 1}}{4} + \frac{{11}}{{15}}} \right).\frac{{2023}}{{2022}} = \left( { - 1 + 1} \right).\frac{{2023}}{{2022}} = 0\].