Đề thi Cuối học kỳ 2 Toán 6 có đáp án (Đề 2)

Tính nhanh: A = 10/3.7 - 5/7.12 - 7/12.19 - 5/19.24

14/14

Tính nhanh: \[A = \frac{{10}}{{3.7}} - \frac{5}{{7.12}} - \frac{7}{{12.19}} - \frac{5}{{19.24}}\]

0/3000 ký tự
Giải thích

Hướng dẫn giải:

\[A = \frac{{10}}{{3.7}} - \frac{5}{{7.12}} - \frac{7}{{12.19}} - \frac{5}{{19.24}}\]

\(A = \frac{{3 + 7}}{{3.7}} - \frac{{12 - 7}}{{7.12}} - \frac{{19 - 12}}{{12.19}} - \frac{{24 - 19}}{{19.24}}\)

\[A = \left( {\frac{3}{{3.7}} + \frac{7}{{3.7}}} \right) - \left( {\frac{{12}}{{7.12}} - \frac{7}{{7.12}}} \right) - \left( {\frac{{19}}{{12.19}} - \frac{{12}}{{12.19}}} \right) - \left( {\frac{{24}}{{19.24}} - \frac{{19}}{{19.24}}} \right)\]

\[A = \left( {\frac{1}{7} + \frac{1}{3}} \right) - \left( {\frac{1}{7} - \frac{1}{{12}}} \right) - \left( {\frac{1}{{12}} - \frac{1}{{19}}} \right) - \left( {\frac{1}{{19}} - \frac{1}{{24}}} \right)\]

\[A = \frac{1}{3} + \frac{1}{7} - \frac{1}{7} + \frac{1}{{12}} - \frac{1}{{12}} + \frac{1}{{19}} - \frac{1}{{19}} + \frac{1}{{24}}\]

\[A = \frac{1}{3} + \left( {\frac{1}{7} - \frac{1}{7}} \right) + \left( {\frac{1}{{12}} - \frac{1}{{12}}} \right) + \left( {\frac{1}{{19}} - \frac{1}{{19}}} \right) + \frac{1}{{24}}\]

\(A = \frac{1}{3} + \frac{1}{{24}}\)

\(A = \frac{8}{{24}} + \frac{1}{{24}}\)

\(A = \frac{9}{{24}}\)

\(A = \frac{3}{8}\)

Vậy \(A = \frac{3}{8}\)