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Ta có : \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}.\frac{2}{3}....\frac{18}{19}.\frac{19}{20}\)
\(=\frac{1.2....18.19}{2.3...19.20}\)
\(=\frac{1}{20}>\frac{1}{21}\)
Vậy A > 1/21
\(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{10}-1\right)\)
\(\Rightarrow A=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-9}{10}\)
\(\Rightarrow A=\frac{-1}{10}\)
Dễ thấy \(\frac{1}{10}< \frac{1}{9}\Rightarrow\frac{-1}{10}>\frac{-1}{9}\)
\(A=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}.....\frac{-9}{10}\)
\(A=\frac{-1}{10}\)
\(\frac{-1}{10}>\frac{-1}{9}\Rightarrow A>\frac{-1}{9}\)
đ/s:..
Cho \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
so sánh B với \(\frac{3}{4}\)
Ta có:\(\frac{1}{2^2}=\frac{1}{4}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
....
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Leftrightarrow B< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{100}\)
B < \(\frac{1}{4}\) < \(\frac{3}{4}\)
\(\Leftrightarrow B< \frac{3}{4}\)
\(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\cdot...\left(\frac{1}{10}-1\right)\)
\(A=\left(\frac{1}{2}-\frac{2}{2}\right)\left(\frac{1}{3}-\frac{3}{3}\right)\cdot...\cdot\left(\frac{1}{10}-\frac{10}{10}\right)\)
\(A=\left(-\frac{1}{2}\right)\cdot\left(-\frac{2}{3}\right)\cdot...\cdot\left(-\frac{9}{10}\right)\)
\(A=\frac{-1}{2}\cdot\frac{-2}{3}\cdot...\cdot\frac{-9}{10}\)
\(A=\frac{\left(-1\right)\cdot\left(-2\right)\cdot...\cdot\left(-9\right)}{2\cdot3\cdot...\cdot10}\)
\(A=\frac{\left(-1\right)\cdot2\cdot...\cdot9}{2\cdot3\cdot...\cdot10}=\frac{-1}{10}\)
Mà \(\frac{-1}{10}>\frac{-1}{9}\)nên A > -1/9
Phần cuối tương tự