Kết quả của phép tính \(0,{6^5}:0,{6^3}\) là

Hướng dẫn giải

Đáp án: \(35\)

Vì \(xy\parallel mn\) nên \(\widehat {xAB} = \widehat {ABn} = 70^\circ \) (so le trong).

Ta có tia \(BC\) là tia phân giác của \(\widehat {ABn}\) nên \(\widehat {ABC} = \widehat {CBn} = \widehat {\frac{{ABn}}{2}} = 35^\circ \).

Vì \(xy\parallel mn\) nên \(\widehat {ACB} = \widehat {CBn} = 35^\circ \).

Hướng dẫn giải

Ta có: \(S = \frac{1}{4} + \frac{2}{{{4^2}}} + \frac{3}{{{4^3}}} + \frac{4}{{{4^4}}} + .... + \frac{{2023}}{{{4^{2023}}}}\)

\(4S = 4\left( {\frac{1}{4} + \frac{2}{{{4^2}}} + \frac{3}{{{4^3}}} + \frac{4}{{{4^4}}} + .... + \frac{{2023}}{{{4^{2023}}}}} \right)\)

\(4S = 1 + \frac{2}{4} + \frac{3}{{{4^2}}} + \frac{4}{{{4^3}}} + \frac{5}{{{4^4}}} + .... + \frac{{2023}}{{{4^{2022}}}}\)

\(4S - S = 1 + \frac{2}{4} + \frac{3}{{{4^2}}} + \frac{4}{{{4^3}}} + \frac{5}{{{4^4}}} + .... + \frac{{2023}}{{{4^{2022}}}} - \left( {\frac{1}{4} + \frac{2}{{{4^2}}} + \frac{3}{{{4^3}}} + \frac{4}{{{4^4}}} + .... + \frac{{2023}}{{{4^{2023}}}}} \right)\)

\(3S = 1 + \frac{2}{4} + \frac{3}{{{4^2}}} + \frac{4}{{{4^3}}} + \frac{5}{{{4^4}}} + .... + \frac{{2023}}{{{4^{2022}}}} - \frac{1}{4} - \frac{2}{{{4^2}}} - \frac{3}{{{4^3}}} - \frac{4}{{{4^4}}} - .... - \frac{{2023}}{{{4^{2023}}}}\)

\(3S = 1 + \left( {\frac{2}{4} - \frac{1}{4}} \right) + \left( {\frac{3}{{{4^2}}} - \frac{2}{{{4^2}}}} \right) + \left( {\frac{4}{{{4^3}}} - \frac{3}{{{4^3}}}} \right) + \left( {\frac{5}{{{4^4}}} - \frac{4}{{{4^4}}}} \right) + .... + \left( {\frac{{2023}}{{{4^{2022}}}} - \frac{{2022}}{{{4^{2022}}}}} \right) - \frac{{2023}}{{{4^{2023}}}}\)

\(3S = 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + \frac{1}{{{4^3}}} + \frac{1}{{{4^4}}} + .... + \frac{1}{{{4^{2022}}}} - \frac{{2023}}{{{4^{2023}}}}\)

Nhận thấy \(3S < 1\).

Đặt \(A = 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + \frac{1}{{{4^3}}} + \frac{1}{{{4^4}}} + .... + \frac{1}{{{4^{2022}}}}\)

\(4A = 4 + 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + \frac{1}{{{4^3}}} + \frac{1}{{{4^4}}} + .... + \frac{1}{{{4^{2021}}}}\)

\(4A - A = 4 + 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + \frac{1}{{{4^3}}} + \frac{1}{{{4^4}}} + .... + \frac{1}{{{4^{2021}}}} - \left( {1 + \frac{1}{4} + \frac{1}{{{4^2}}} + \frac{1}{{{4^3}}} + \frac{1}{{{4^4}}} + .... + \frac{1}{{{4^{2022}}}}} \right)\)

\(3A = 4 - \frac{1}{{{4^{2022}}}}\)

Nhận thấy \(4 - \frac{1}{{{4^{2022}}}} < 4\) hay \(3A < 4\) suy ra \(A < \frac{4}{3}\).

Do đó, \(3S < A\) nên \(S < \frac{A}{3}\) hay \(S < \frac{4}{9} < \frac{4}{8} = \frac{1}{2}.\)

Vậy \(S < \frac{1}{2}.\)