Câu hỏi:

29/06/2025 7

Chứng tỏ rằng \(\frac{1}{{11}} + \frac{1}{{12}} + \frac{1}{{13}} + ... + \frac{1}{{70}} > \frac{4}{3}\).

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Trả lời:

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Giải bởi Vietjack

Đặt \(A = \frac{1}{{11}} + \frac{1}{{12}} + \frac{1}{{13}} + ... + \frac{1}{{70}}\)

Ta có: \(A = \frac{1}{{11}} + \frac{1}{{12}} + \frac{1}{{13}} + ... + \frac{1}{{30}} + \frac{1}{{31}} + \frac{1}{{32}} + ... + \frac{1}{{50}} + \frac{1}{{51}} + \frac{1}{{52}} + ... + \frac{1}{{70}}\)

\(A = \left( {\frac{1}{{11}} + \frac{1}{{12}} + \frac{1}{{13}} + ... + \frac{1}{{30}}} \right) + \left( {\frac{1}{{31}} + \frac{1}{{32}} + ... + \frac{1}{{50}}} \right) + \left( {\frac{1}{{51}} + \frac{1}{{52}} + ... + \frac{1}{{70}}} \right)\)

Nhận thấy \(\frac{1}{{11}} + \frac{1}{{12}} + \frac{1}{{13}} + ... + \frac{1}{{30}} > \frac{1}{{30}} + \frac{1}{{30}} + .... + \frac{1}{{30}}\) hay \(\frac{1}{{11}} + \frac{1}{{12}} + \frac{1}{{13}} + ... + \frac{1}{{30}} > \frac{1}{{30}}.20 = \frac{2}{3}\)

                  \(\frac{1}{{31}} + \frac{1}{{32}} + ... + \frac{1}{{50}} > \frac{1}{{50}} + \frac{1}{{50}} + .... + \frac{1}{{50}}\) hay \(\frac{1}{{31}} + \frac{1}{{32}} + ... + \frac{1}{{50}} > \frac{1}{{50}}.20 = \frac{2}{5}\).

                  \(\frac{1}{{51}} + \frac{1}{{52}} + ... + \frac{1}{{70}} > \frac{1}{{70}} + \frac{1}{{70}} + .... + \frac{1}{{70}}\) hay \(\frac{1}{{51}} + \frac{1}{{52}} + ... + \frac{1}{{70}} > \frac{1}{{70}}.20 = \frac{2}{7}\).

Do đó, \(A > \frac{2}{3} + \frac{2}{5} + \frac{2}{7}\) hay \(A > \frac{{142}}{{105}} > \frac{{140}}{{105}} = \frac{4}{3}\).

Vậy \(A > \frac{4}{3}\) (đpcm)

CÂU HỎI HOT CÙNG CHỦ ĐỀ

Lời giải

a) \(\frac{5}{3}.\frac{{ - 8}}{{25}} + \frac{5}{3}.\frac{{ - 21}}{{25}} + \frac{5}{3}.\frac{4}{{25}} - \frac{3}{5}.\)

\( = \frac{5}{3}.\left( {\frac{{ - 8}}{{25}} + \frac{{ - 21}}{{25}} + \frac{4}{{25}}} \right) - \frac{3}{5}\)

\( = \frac{5}{3}.\left( {\frac{{ - 25}}{{25}}} \right) - \frac{3}{5}\)

\( = \frac{5}{3}.\left( { - 1} \right) - \frac{3}{5}\)

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

\( = \frac{{ - 25}}{{15}} - \frac{9}{{15}}\)

\( = \frac{{ - 34}}{{15}}.\)

b) \(\frac{6}{{13}}.\frac{{ - 11}}{{21}} + \frac{6}{{13}}.\frac{{ - 15}}{{21}} + \frac{6}{{13}}.\frac{5}{{21}} - \frac{{13}}{6}\)

\( = \frac{6}{{13}}.\left( {\frac{{ - 11}}{{21}} + \frac{{ - 15}}{{21}} + \frac{5}{{21}}} \right) - \frac{{13}}{6}\)

\( = \frac{6}{{13}}.\left( {\frac{{ - 21}}{{21}}} \right) - \frac{{13}}{6}\)

\( = \frac{6}{{13}}.\left( { - 1} \right) - \frac{{13}}{6}\)

\( = \frac{{ - 6}}{{13}} - \frac{{13}}{6}\)

\( = \frac{{ - 36}}{{78}} - \frac{{169}}{{78}}\)

\( = \frac{{ - 205}}{{78}}\).

Lời giải

Đáp án: \( - 12\)

Ta có: \(\frac{x}{{12}} + \frac{1}{3} = \frac{{ - 2}}{3}\)

\(\frac{x}{{12}} = \frac{{ - 2}}{3} - \frac{1}{3}\)

\(\frac{x}{{12}} = - 1\)

\(x = - 1.12\)

\(x = - 12\)

Vậy \(x = - 12.\)