Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt :\(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}\)
\(N=\frac{2}{3}.\frac{4}{5}...\frac{10000}{10001}\)
Ta thấy:\(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};....;\frac{9999}{10000}< \frac{10000}{10001}\)
Mặt khác ta thấy:
\(C.N=\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\right).\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{10000}{10001}\right)\)
\(C.N=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{9999}{10000}.\frac{10000}{10001}\)
\(C.N=\frac{1.2.3....9999.10000}{2.3.4....10000.10001}\)
Rút gọn phép tính \(C.N\)
\(C.N=\frac{1}{10001}\)
\(C.C< N\Rightarrow C.C< C.N\)
Hay\(C.C< \frac{1}{10001}< \frac{1}{10000}=\frac{1}{10}.\frac{1}{10}\)
\(\Rightarrow C< \frac{1}{10000}\)(đpcm)
Ta có:
\(C=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\cdot\cdot\cdot\cdot\frac{9999}{10000}\)
Đặt \(I=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\cdot\cdot\cdot\cdot\frac{10000}{10001}\)
Ta có: \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};.....;\frac{9999}{10000}< \frac{10000}{10001}\)
\(\Rightarrow C< D\)
Lại có: \(C\cdot D=\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\cdot\cdot\cdot\cdot\frac{9999}{10000}\right)\cdot\left(\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\cdot\cdot\cdot\cdot\frac{10000}{10001}\right)\)
\(\Leftrightarrow C\cdot D=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\cdot\cdot\cdot\cdot\frac{9999}{10000}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\cdot\cdot\cdot\cdot\frac{10000}{10001}\)
\(\Leftrightarrow C\cdot D=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\cdot\cdot\cdot\cdot\frac{9999}{10000}\cdot\frac{10000}{10001}\)
\(\Leftrightarrow C\cdot D=\frac{1}{10001}\)
Mà C<D \(\Rightarrow C\cdot C< C\cdot D\)
Hay \(C\cdot C< \frac{1}{10001}\)
\(\Rightarrow C< \frac{1}{10001}< \frac{1}{100}\)
Vậy \(C< \frac{1}{100}\left(đpcm\right)\)
\(S=\frac{1}{4}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
\(S< \frac{1}{4}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)\)
\(S< \frac{1}{4}\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(S< \frac{1}{4}\left(1+1-\frac{1}{50}\right)\)
\(S< \frac{1}{4}.\frac{99}{50}=\frac{99}{200}< \frac{1}{2}\)
VẬY\(S< \frac{1}{2}\)
a)\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
\(=\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{8}-\frac{1}{16}\right)+\left(\frac{1}{32}-\frac{1}{64}\right)\)
\(=\frac{1}{4}+\frac{1}{16}+\frac{1}{64}\)
\(=\frac{16+4+1}{64}\)
\(=\frac{21}{64}< \frac{1}{3}\)(đpcm)
khó ngược
làm nhé Thánh