Đề thi Giữa kì 2 Toán 6 trường THCS Lê Quý Đôn (Hồ Chí Minh) năm 2024-2025 có đáp án

Thực hiện phép tính: (a) 5/4 − − 2/3 + 7/− 12 .

Giải thích

a) \(\frac{5}{4} - \,\frac{{ - 2}}{3} + \,\frac{7}{{ - 12}}\,\)

\( = \frac{{15}}{{12}} - \,\frac{{ - 8}}{{12}} + \,\frac{{ - 7}}{{12}}\,\)

\( = \frac{{16}}{{12}} = \frac{4}{3}.\)

b) \(\left( {\frac{3}{5} + \,\frac{{ - 4}}{7}} \right):\,\frac{{2025}}{{2024}}\, + \left( {\frac{{ - 3}}{7} + \,\frac{2}{5}} \right):\,\frac{{2025}}{{2024}}\,\)

\[ = \left( {\frac{3}{5} + \,\frac{{ - 4}}{7}} \right) \cdot \frac{{2024}}{{2025}}\, + \left( {\frac{{ - 3}}{7} + \,\frac{2}{5}} \right) \cdot \frac{{2024}}{{2025}}\,\]

\[ = \left( {\frac{3}{5} + \,\frac{{ - 4}}{7} + \frac{{ - 3}}{7} + \,\frac{2}{5}} \right) \cdot \frac{{2024}}{{2025}}\,\]

\[ = \left[ {\left( {\frac{3}{5} + \,\frac{2}{5}} \right) + \left( {\frac{{ - 4}}{7} + \frac{{ - 3}}{7}} \right)} \right] \cdot \frac{{2024}}{{2025}}\,\]

\[ = \left[ {\frac{5}{5} + \frac{{ - 7}}{7}} \right] \cdot \frac{{2024}}{{2025}}\,\]

\[ = \left[ {1 + \left( { - 1} \right)} \right] \cdot \frac{{2024}}{{2025}}\,\]\[ = 0 \cdot \frac{{2024}}{{2025}}\, = 0.\]