a) Tính bằng cách thuận tiện: (252525 ´ 262626 – 505050 ´ 131313) : 2526

= (25 ´ 10101 ´ 26 ´ 10101 – 50 ´ 10101 ´ 13 ´ 10101) : 2526

= (25 ´ 10101 ´ 2 ´ 13 ´ 10101 – 25 ´ 2 ´ 10101 ´ 13 ´ 10101) : 2526

= 0 : 2526

= 0

b) \(A = \frac{1}{2} + \frac{5}{6} + \frac{{11}}{{12}} + \frac{{19}}{{20}} + \frac{{41}}{{42}} + \frac{{55}}{{56}} + \frac{{71}}{{72}}\)

=\(\left( {1 - \frac{1}{2}} \right) + \left( {1 - \frac{1}{6}} \right) + \left( {1 - \frac{1}{{12}}} \right) + \left( {1 - \frac{1}{{20}}} \right) + \left( {1 - \frac{1}{{30}}} \right) + \left( {1 - \frac{1}{{42}}} \right) + \left( {1 - \frac{1}{{56}}} \right) + \left( {1 - \frac{1}{{72}}} \right)\)

=\((1 + 1 + 1 + 1 + 1 + 1 + 1 + 1) - \left( {\frac{1}{2} + \frac{1}{6} + \frac{1}{{12}} + \frac{1}{{20}} + \frac{1}{{30}} + \frac{1}{{42}} + \frac{1}{{56}} + \frac{1}{{72}}} \right)\)

=\(8 - \left( {\frac{1}{{1 \times 2}} + \frac{1}{{2 \times 3}} + \frac{1}{{3 \times 4}} + \frac{1}{{4 \times 5}} + \frac{1}{{5 \times 6}} + \frac{1}{{6 \times 7}} + \frac{1}{{7 \times 8}} + \frac{1}{{8 \times 9}}} \right)\)

=\(8 - \left( {1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \frac{1}{5} - \frac{1}{6} + \frac{1}{6} - \frac{1}{7} + \frac{1}{7} - \frac{1}{8} + \frac{1}{8} - \frac{1}{9}} \right)\)

=\(8 - \left( {1 - \frac{1}{9}} \right)\)

=\(8 - 1 + \frac{1}{9}\)

=\(7 + \frac{1}{9} > 7\)

Có 5 đôi giày màu đỏ và 7 đôi giày màu xanh cùng cỡ để để trong một hộp kín.