Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

Hình như sửa đề lại nhé
Câu hỏi của Tuấn Anh - Toán lớp 7 - Học toán với OnlineMath
Tham khảo nhé

Câu hỏi của Vũ Thị Kim Oanh - Toán lớp 7 - Học toán với OnlineMath
Tham khảo

\(A=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+....+\frac{10}{5^{10}}+\frac{11}{5^{11}}\)
\(\Rightarrow5A=1+\frac{2}{5}+\frac{3}{5^2}+....+\frac{10}{5^9}+\frac{11}{5^{10}}\)
\(\Rightarrow5A-A=\left(1+\frac{2}{5}+...+\frac{11}{5^{10}}\right)-\left(\frac{1}{5}+\frac{2}{5^2}+...+\frac{11}{5^{11}}\right)\)
\(\Rightarrow4A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}-\frac{11}{5^{11}}\)(1)
Đặt \(B=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}\)
\(\Rightarrow5B=5+1+\frac{1}{5}+...+\frac{1}{5^9}\)
\(\Rightarrow5B-B=\left(5+1+...+\frac{1}{5^9}\right)-\left(1+\frac{1}{5}+...+\frac{1}{5^{10}}\right)\)
\(\Rightarrow4B=5-\frac{1}{5^{10}}< 5\)
\(\Rightarrow B< \frac{5}{4}\)(2)
Thay (2) vào (1) \(\Rightarrow4A< \frac{5}{4}-\frac{11}{5^{11}}< \frac{5}{4}\)
\(\Rightarrow A< \frac{5}{16}\left(đpcm\right)\)

A= 1-\(\frac{1}{2}\) +\(\frac{1}{3}\) - \(\frac{1}{4}\) +\(\frac{1}{5}\)- \(\frac{1}{6}\) + ...+ \(\frac{1}{99}\) - \(\frac{1}{100}\)
= 1+ \(\frac{1}{2}\) + \(\frac{1}{3}\) + \(\frac{1}{4}\) + \(\frac{1}{5}\) + \(\frac{1}{6}\) + ...+ \(\frac{1}{99}\) + \(\frac{1}{100}\) - 2 ( \(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{6}\) + ...+ \(\frac{1}{100}\) )
= 1+ \(\frac{1}{2}\) + \(\frac{1}{3}\) + \(\frac{1}{4}\) + ...+ \(\frac{1}{99}\) + \(\frac{1}{100}\)
= \(\frac{1}{51}\) + \(\frac{1}{52}\) +...+ \(\frac{1}{100}\)
= (\(\frac{1}{51}\) + \(\frac{1}{52}\) + ... + \(\frac{1}{75}\) ) + ( \(\frac{1}{76}\) + \(\frac{1}{77}\) + ... + \(\frac{1}{100}\) )
Ta có : \(\frac{1}{51}\) > \(\frac{1}{52}\) > \(\frac{1}{53}\) > ... > \(\frac{1}{75}\)
\(\frac{1}{76}\) > \(\frac{1}{77}\) > \(\frac{1}{78}\) > ... > \(\frac{1}{100}\)
=> A > \(\frac{1}{75}.25\) + \(\frac{1}{100}.25\) = \(\frac{1}{3}\) + \(\frac{1}{4}\) = \(\frac{7}{12}\)
=> A< \(\frac{1}{51}.25\) + \(\frac{1}{75}.25\) < \(\frac{1}{50}.25\) + \(\frac{1}{75}.25\) = \(\frac{1}{2}\) + \(\frac{1}{3}\) = \(\frac{5}{6}\)
Vậy \(\frac{7}{12}\) < A < \(\frac{5}{6}\)
Tick nha