Bộ 5 đề thi cuối kì 2 Toán 6 Kết nối tri thức (Tự luận) có đáp án - Đề 4

Thực hiện phép tính (tính hợp lí nếu có thể): a) 2/3 - (- 5/7 + 2/3)

Giải thích

a) \(\frac{2}{3} - \left( {\frac{{ - 5}}{7} + \frac{2}{3}} \right)\)\( = \frac{2}{3} + \frac{5}{7} - \frac{2}{3}\)

\( = \left( {\frac{2}{3} - \frac{2}{3}} \right) + \frac{5}{7}\)\( = 0 + \frac{5}{7}\)\( = \frac{5}{7}.\)

c) \(\frac{2}{3}:\left( {\frac{2}{5} + \frac{1}{2}} \right) + \frac{2}{3}:\left( {\frac{1}{4} - \frac{4}{7}} \right)\)

\( = \frac{2}{3}:\frac{9}{{10}} + \frac{2}{3}:\frac{{ - 9}}{{28}}\)

\( = \frac{2}{3} \cdot \frac{{10}}{9} + \frac{2}{3} \cdot \frac{{ - 28}}{9}\)

\[ = \frac{2}{3} \cdot \left( {\frac{{10}}{9} + \frac{{ - 28}}{9}} \right)\]

\( = \frac{2}{3} \cdot \left( { - 2} \right) = \frac{{ - 4}}{3}\).

b) \(2,35:\left( { - 0,01} \right) + 650 \cdot \left( { - 0,1} \right)\)

\( = - 235 - 65\)

\( = - 300.\)

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

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

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

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

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

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