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tử là M mẫu là N ta dc
\(M=2008+\frac{2007}{2}+...+\frac{1}{2008}\)
\(=\left(1+...+1\right)+\frac{2007}{2}+...+\frac{1}{2008}\)
\(=\frac{2009}{2}+...+\frac{2009}{2008}+\frac{2009}{2009}\)
\(=2009\left(\frac{1}{2}+...+\frac{1}{2008}+\frac{1}{2009}\right)\)
vậy ta có
\(A=\frac{M}{N}=\frac{2009\left(\frac{1}{2}+...+\frac{1}{2008}+\frac{1}{2009}\right)}{\frac{1}{2}+...+\frac{1}{2008}+\frac{1}{2009}}\)\(=2009\)
5A=\(\frac{1}{5}+\frac{2}{5^2}...+\frac{n}{5^n}...+\frac{11}{5^{11}}\)
=>4A=5A-A=\(\frac{1}{5}+\frac{1}{5^2}...+\frac{1}{5^{11}}-\frac{11}{5^{12}}\)
=>20A=\(1+\frac{1}{5}+...+\frac{1}{5^{10}}-\frac{11}{5^{11}}\)
=>16A=20A-4A=\(1-\frac{1}{5^{11}}+\frac{11}{5^{12}}-\frac{11}{5^{11}}\)
Mà \(1-\frac{1}{5^{11}}< 1\),\(\frac{11}{5^{12}}-\frac{11}{5^{11}}< 0\)
=>16A<1
Do đó: A<1/16(đpcm)
\(\frac{-5}{3}-\left(\frac{4}{5}-\frac{1}{2}\right)-\left|\frac{3}{4}-\frac{5}{2}+\frac{1}{3}\right|\)
\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\left|\frac{3}{4}+\frac{-5}{2}+\frac{1}{3}\right|\)
\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\left(\frac{3}{4}+\frac{-5}{2}+\frac{1}{3}\right)\)
\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\frac{3}{4}+\frac{5}{2}-\frac{1}{3}\)
\(=\left(\frac{-5}{3}-\frac{1}{3}\right)+\left(\frac{1}{2}+\frac{5}{2}\right)-\left(\frac{4}{5}+\frac{3}{4}\right)\)
\(=\frac{-6}{3}+\frac{6}{2}-\left(\frac{16}{20}+\frac{15}{20}\right)\)
\(=-2+3-\frac{31}{20}\)
\(=1-\frac{31}{20}=\frac{-11}{20}\)
\(\frac{-5}{3}-\left(\frac{4}{5}-\frac{1}{2}\right)-\left|\frac{3}{4}-\frac{5}{2}+\frac{1}{3}\right|\)
\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\left|\frac{3}{4}+\frac{-5}{2}+\frac{1}{3}\right|\)
\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\left(\frac{3}{4}+\frac{-5}{2}+\frac{1}{3}\right)\)
\(=\frac{-5}{3}-\frac{4}{5}+\frac{1}{2}-\frac{3}{4}+\frac{5}{2}-\frac{1}{3}\)
\(=\left(\frac{-5}{3}-\frac{1}{3}\right)+\left(\frac{1}{2}+\frac{5}{2}\right)-\left(\frac{4}{5}+\frac{3}{4}\right)\)
\(=\frac{-6}{3}+\frac{6}{2}-\left(\frac{16}{20}+\frac{15}{20}\right)\)
\(=\frac{-6}{3}+\frac{6}{2}-\left(\frac{16}{20}+\frac{15}{20}\right)\)
\(=1-\frac{31}{20}=\frac{-11}{20}\)
Ta có: \(A=\frac{2008+\frac{2007}{2}+\frac{2006}{3}+....+\frac{2}{2007}+\frac{1}{2008}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2008}+\frac{1}{2009}}\)
Xét tử : \(2008+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)
\(=\left(1+1+...+1\right)+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)( có 2008 số hạng 1 )
\(=\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+...+\left(1+\frac{2}{2007}\right)+\left(1+\frac{1}{2008}\right)+1\)
\(=\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2007}+\frac{2009}{2008}+\frac{2009}{2009}\)
\(=2009\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)\)
Ghép tử và mẫu....
Vậy A = 2009
Bạn tham khảo ở link này nhé :
Câu hỏi của Tăng Minh Châu - Toán lớp 6 | Học trực tuyến
Đặt A=1/51+1/52+1/53+...+1/52008
5A=1+1/51+1/52+...+1/52007
5A-A=(1+1/51+1/52+...+1/52007)-(1/51+1/52+1/53+...+1/52008)
4A=1-1/52008<1
A<1/4