\(B=\frac{1}{4}+\frac{1}{5}+....+\frac{1}{19}=?\)
\(B=1-\frac{1}{19}\)
\(B=\frac{18}{19}\)
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CTR: \(\frac{1}{5}<\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-...+\frac{1}{98}-\frac{1}{99}<\frac{2}{5}\)
CTR: \(\frac{1}{5}<\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-...+\frac{1}{98}-\frac{1}{99}<\frac{2}{5}\)
CTR
B = \(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}<6\)
B=\(-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
B=\(-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
cho B = .\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+.....+\frac{1}{19}\) Hay chung to B< 1
cho \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2003^2}\)
CTR \(\frac{1}{3}< A< 1\)
Bài 5: Cho B = \(\frac{1}{4} +\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\). Hãy chứng tỏ B > 1.
Cho \(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{99}{100}\)
CTR: \(\frac{1}{15}< A< \frac{1}{10}\)