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\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{98}{99}.\frac{99}{100}\)
\(\Rightarrow A^2>\frac{1}{100}=\frac{1}{10^2}\)
Vậy \(A>\frac{1}{10}\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{9998}{9999}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{9998}{9999}.\frac{9999}{10000}\)
\(\Rightarrow A^2>\frac{1}{10000}=\frac{1}{100^2}\)
\(VayA>\frac{1}{100}=B\)
Sửa N=\(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{100}{101}\)
Ta có : \(\frac{1}{2}< \frac{2}{3}\); \(\frac{3}{4}< \frac{4}{5}\); \(\frac{5}{6}< \frac{6}{7}\); ... ; \(\frac{99}{100}< \frac{100}{101}\)
\(\Rightarrow\)\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)hay M < N
b) M .N = \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}.\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}=\frac{1.2.3.4.5.6...99.100}{2.3.4.5.6.7...100.101}=\frac{1}{101}\)
c) vì M < N nên M. M < M . N = \(\frac{1}{101}\)\(< \frac{1}{100}\)
\(\Rightarrow M< \frac{1}{10}\)
\(A=\frac{1}{2}\times\frac{3}{4}......\frac{9999}{10000}\)
Đặt : \(B=\frac{2}{3}\times\frac{4}{5}\times\frac{6}{7}.......\frac{10000}{10001}\)
Vì \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};.....\frac{9999}{10000}< \frac{10000}{10001}\)
Nên A<B mà A>0; B>0
\(\Rightarrow A^2< A\times B=\left(\frac{1}{2}\times\frac{3}{4}\times\frac{5}{6}.....\frac{9999}{10000}\right)\times\left(\frac{2}{3}\times\frac{4}{5}\times\frac{6}{7}......\frac{10000}{10001}\right)\)\(=\frac{1}{2}\times\frac{2}{3}\times\frac{4}{5}......\frac{9999}{10000}\times\frac{10000}{10001}\)\(=\frac{1}{10001}< \frac{1}{10000}=\frac{1}{100^2}=0.01^2\)\(\Rightarrow A^2< 0.01^2\)hay A < 0.01
\(A=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{99!}+\frac{1}{100!}\)
\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(A< 1+1-\frac{1}{100}\)
\(A< 2-\frac{1}{100}< 2\)
\(A=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{99!}+\frac{1}{100!}\)
\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(A< 1+1-\frac{1}{100}\)
\(A< 2-\frac{1}{100}< 2\)
1/100 hả e hay là 1/10
Dạ 1/100