\[A = \frac{1}{{11}} + \frac{1}{{12}} + \frac{1}{{13}} + ... + \frac{1}{{70}}\]

\[A = \left( {\frac{1}{{11}} + \frac{1}{{12}} + ... + \frac{1}{{20}}} \right) + \left( {\frac{1}{{21}} + \frac{1}{{22}} + ... + \frac{1}{{30}}} \right) + \left( {\frac{1}{{31}} + \frac{1}{{32}} + ... + \frac{1}{{40}}} \right)\]

\[ + \left( {\frac{1}{{41}} + \frac{1}{{42}} + ... + \frac{1}{{50}}} \right) + \left( {\frac{1}{{51}} + \frac{1}{{52}} + ... + \frac{1}{{60}}} \right) + \left( {\frac{1}{{61}} + \frac{1}{{62}} + ... + \frac{1}{{70}}} \right)\]

⦁ Ta có:

\[\frac{1}{{11}} < \frac{1}{{10}};\frac{1}{{12}} < \frac{1}{{10}};...;\frac{1}{{20}} < \frac{1}{{10}}\]

\[\frac{1}{{21}} < \frac{1}{{20}};\frac{1}{{22}} < \frac{1}{{20}};...;\frac{1}{{30}} < \frac{1}{{20}}\]

....

\[\frac{1}{{61}} < \frac{1}{{60}};\frac{1}{{62}} < \frac{1}{{60}};...;\frac{1}{{70}} < \frac{1}{{60}}\]

Do đó

\[A < \frac{1}{{10}} \cdot 10 + \frac{1}{{20}} \cdot 10 + \frac{1}{{30}} \cdot 10 + \frac{1}{{40}} \cdot 10 + \frac{1}{{50}} \cdot 10 + \frac{1}{{60}} \cdot 10\]

\[A < 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}\]

\[A < 1 + \frac{{29}}{{20}}\]

A < 2,45 hay A < 2,5 (1)

⦁ Ta có:

\[\frac{1}{{11}} > \frac{1}{{20}};\frac{1}{{12}} > \frac{1}{{20}};...;\frac{1}{{19}} > \frac{1}{{20}}\]

\[\frac{1}{{21}} > \frac{1}{{30}};\frac{1}{{22}} > \frac{1}{{30}};...;\frac{1}{{29}} > \frac{1}{{30}}\]

\[\frac{1}{{31}} > \frac{1}{{60}};\frac{1}{{32}} > \frac{1}{{60}};...;\frac{1}{{59}} > \frac{1}{{60}}\]

Suy ra \[A > \frac{1}{{20}} \cdot 10 + \frac{1}{{30}} \cdot 10 + \frac{1}{{60}} \cdot 30 + \left( {\frac{1}{{60}} + \frac{1}{{61}} + ... + \frac{1}{{70}}} \right)\]

\[A > \frac{1}{2} + \frac{1}{3} + \frac{1}{2} + \left( {\frac{1}{{60}} + \frac{1}{{61}} + ... + \frac{1}{{70}}} \right)\]

\[A > \frac{4}{3} + \left( {\frac{1}{{60}} + \frac{1}{{61}} + ... + \frac{1}{{70}}} \right) > \frac{4}{3}.\] (2)

Từ (1) và (2) suy ra \[\frac{4}{3} < A < 2,5.\]

Tìm x để 10 chia hết cho x ‒ 7?