Bộ 5 đề thi cuối kì 2 Toán 6 Chân trời sáng tạo (Tự luận) có đáp án - Đề 3

Thực hiện phép tính (tính hợp lí nếu có thể): a) \(\frac{2}{5} - \frac{1}{5} \cdot \frac{3}{{ - 4}}.\)                                 b) \(\left( { - 12,5} \right) + 17,55 + \left( { - 3,5

Giải thích

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

\( = \frac{2}{5} - \frac{3}{{ - 20}}\)

\( = \frac{8}{{20}} + \frac{3}{{20}}\)

\( = \frac{{11}}{{20}}.\)

b) \[\left( { - 12,5} \right) + 17,55 + \left( { - 3,5} \right) - \left( { - 2,45} \right)\]

\[ = \left[ {\left( { - 12,5} \right) + \left( { - 3,5} \right)} \right] + \left[ {17,55 - \left( { - 2,45} \right)} \right]\]

\[ = \left( { - 16} \right) + \left[ {17,55 + 2,45} \right]\]

\[ = \left( { - 16} \right) + 20\]

\[ = 4.\]

c) \(\frac{2}{3}:\frac{4}{5} - \frac{5}{4} + \frac{1}{3}:\frac{4}{5}\)

\( = \frac{2}{3} \cdot \frac{5}{4} - \frac{5}{4} + \frac{1}{3} \cdot \frac{5}{4}\)

\( = \frac{5}{4} \cdot \left( {\frac{2}{3} - 1 + \frac{1}{3}} \right) = \frac{5}{4} \cdot \left[ {\left( {\frac{2}{3} + \frac{1}{3}} \right) - 1} \right]\)

\( = \frac{5}{4} \cdot \left[ {\frac{3}{3} - 1} \right] = \frac{5}{4} \cdot \left( {1 - 1} \right)\)

\( = \frac{5}{4} \cdot 0 = 0.\)

d) \[1\frac{{13}}{{15}} \cdot {\left( {0,5} \right)^2} \cdot 3 + \left( {40\%  - 1\frac{{19}}{{60}}} \right):1\frac{7}{8}\]

\( = \frac{{28}}{{15}} \cdot {\left( {\frac{1}{2}} \right)^2} \cdot 3 + \left( {\frac{2}{5} - \frac{{79}}{{60}}} \right):\frac{{15}}{8}\)

\( = \frac{{28}}{{15}} \cdot \frac{1}{4} \cdot 3 + \left( {\frac{{24}}{{60}} - \frac{{79}}{{60}}} \right) \cdot \frac{8}{{15}}\)

\[ = \frac{7}{5} + \frac{{ - 55}}{{60}} \cdot \frac{8}{{15}}\]

\[ = \frac{7}{5} + \frac{{ - 22}}{{45}}\]

\[ = \frac{{63}}{{45}} + \frac{{ - 22}}{{45}} = \frac{{41}}{{43}}.\]