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![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right).....\left(1-\frac{1}{100}\right)\)
\(=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...............\frac{99}{100}\)
\(=\frac{3.8.15......99}{4.9.16....100}=\frac{\left(1.3\right).\left(2.4\right).\left(3.5\right).......\left(9.11\right)}{\left(2.2\right).\left(3.3\right).\left(4.4\right)......\left(10.10\right)}\)
\(=\frac{\left(1.2.3.....9\right).\left(3.4.5......11\right)}{\left(2.3.4.....10\right).\left(2.3.4.......10\right)}=\frac{1}{10}.\frac{11}{2}=\frac{11}{20}< \frac{11}{21}\)
Vậy B<11/21
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)...\left(1-\dfrac{1}{81}\right)\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}...\dfrac{99}{100}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}...\dfrac{9.11}{10.10}=\left(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{9}{10}\right).\left(\dfrac{3}{2}.\dfrac{4}{3}...\dfrac{11}{10}\right)=\dfrac{1}{10}.\dfrac{11}{2}=\dfrac{11}{20}>\dfrac{11}{21}\)
\(B=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)...\left(1-\dfrac{1}{9}\right)\left(1+\dfrac{1}{9}\right)\left(1-\dfrac{1}{10}\right)\left(1+\dfrac{1}{10}\right)\\ B=\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{8}{9}\cdot\dfrac{9}{10}\right)\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{10}{9}\cdot\dfrac{11}{10}\right)\\ B=\dfrac{1}{10}\cdot\dfrac{11}{2}=\dfrac{11}{20}>\dfrac{11}{21}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{18}{19}.\frac{19}{20}\)
\(A=\frac{1}{20}\)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)........\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(\Leftrightarrow A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...........\frac{18}{19}.\frac{19}{20}\)
\(\Leftrightarrow A=\frac{1}{20}>\frac{1}{21}\)
\(\Leftrightarrow A>\frac{1}{21}\)
\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)................\left(1-\frac{1}{100}\right)\)
\(\Leftrightarrow B=\frac{3}{4}.\frac{8}{9}..................\frac{99}{100}\)
\(B=\frac{1.3}{2^2}.\frac{2.4}{3^2}................\frac{9.11}{50^2}\)
\(B=\frac{11}{50}< \frac{11}{21}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{81}\right).\left(1-\frac{1}{100}\right)\)
\(B=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{80}{81}.\frac{99}{100}\)
\(B=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{8.10}{9.9}.\frac{9.11}{10.10}\)
\(B=\frac{1.2.3...8.9}{2.3.4...9.10}.\frac{3.4.5...10.11}{2.3.4...9.10}\)
\(B=\frac{1}{10}.\frac{11}{2}\)
\(B=\frac{11}{20}>\frac{11}{21}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(\dfrac{1}{4}-1\right).\left(\dfrac{1}{9}-1\right)....\left(\dfrac{1}{100}-1\right).\)
\(\Rightarrow A=\left(-\dfrac{3}{4}\right).\left(-\dfrac{8}{9}\right)....\left(-\dfrac{99}{100}\right)\)
mà A có 9 dấu - \(\left(4;9;16;25;36;49;64;81;100\right)\)
\(\Rightarrow0>A=\left(-\dfrac{3}{4}\right).\left(-\dfrac{8}{9}\right)....\left(-\dfrac{99}{100}\right)=-\dfrac{1}{2}\)
Ta lại có \(\left\{{}\begin{matrix}\dfrac{1}{2}=\dfrac{21}{42}\\\dfrac{11}{21}=\dfrac{22}{42}\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{2}< \dfrac{11}{21}\Rightarrow-\dfrac{1}{2}>-\dfrac{11}{21}\)
\(\Rightarrow A>-\dfrac{11}{21}\)
\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)...\left(\dfrac{1}{100}-1\right)\)
\(A=\left(-\dfrac{2^2-1}{2^2}\right)\left(-\dfrac{3^2-1}{3^2}\right)...\left(-\dfrac{10^2-1}{10^2}\right)\)
\(A=\left[-\dfrac{1\cdot3}{2\cdot2}\right]\left[-\dfrac{2\cdot4}{3\cdot3}\right]...\left[-\dfrac{9\cdot11}{10\cdot10}\right]\)
Dễ thấy A có 9 thừa số, suy ra
\(A=-\dfrac{1\cdot3\cdot2\cdot4\cdot...\cdot9\cdot11}{2\cdot2\cdot3\cdot3\cdot...\cdot10.10}=-\dfrac{1\cdot11}{2\cdot10}=\dfrac{-11}{20}\)
Vì 20 < 21 nên \(\dfrac{11}{20}>\dfrac{11}{21}\), suy ra \(\dfrac{-11}{20}< \dfrac{-11}{21}\)
Vậy \(A< \dfrac{-11}{21}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\cdot...\left(\frac{1}{10}-1\right)\)
\(A=\left(\frac{1}{2}-\frac{2}{2}\right)\left(\frac{1}{3}-\frac{3}{3}\right)\cdot...\cdot\left(\frac{1}{10}-\frac{10}{10}\right)\)
\(A=\left(-\frac{1}{2}\right)\cdot\left(-\frac{2}{3}\right)\cdot...\cdot\left(-\frac{9}{10}\right)\)
\(A=\frac{-1}{2}\cdot\frac{-2}{3}\cdot...\cdot\frac{-9}{10}\)
\(A=\frac{\left(-1\right)\cdot\left(-2\right)\cdot...\cdot\left(-9\right)}{2\cdot3\cdot...\cdot10}\)
\(A=\frac{\left(-1\right)\cdot2\cdot...\cdot9}{2\cdot3\cdot...\cdot10}=\frac{-1}{10}\)
Mà \(\frac{-1}{10}>\frac{-1}{9}\)nên A > -1/9
Phần cuối tương tự
\(B=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)...\left(\frac{1}{100}-1\right)=\frac{-3}{4}.\frac{-8}{9}...\frac{-99}{100}=-\frac{3.8...99}{4.9....100}\)
\(=-\frac{1.3.2.4...9.11}{2.2.3.3....10.10}=-\frac{\left(1.2...9\right).\left(3.4...11\right)}{\left(2.3...10\right).\left(2.3...10\right)}=-\frac{1.11}{10.2}=-\frac{11}{20}< -\frac{11}{21}\)
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