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![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A= ( 1-\(\frac{1}{4}\)). ( 1-\(\frac{1}{9}\)).( 1-\(\frac{1}{16}\))......(1-\(\frac{1}{100}\))
Ta có:A= ( 1-\(\frac{1}{4}\)). ( 1-\(\frac{1}{9}\)).( 1-\(\frac{1}{16}\))......(1-\(\frac{1}{100}\))
A = \(\frac{3}{4}\).\(\frac{8}{9}\).\(\frac{15}{16}\).......\(\frac{99}{100}\)
A= \(\frac{1.3}{2.2}\). \(\frac{2.4}{3.3}\).\(\frac{3.5}{4.4}\).......\(\frac{9.11}{10.10}\)
A=\(\frac{1.2.3....9}{2.3.4....10}\).\(\frac{3.4.5....11}{2.3.4....10}\)
A= \(\frac{1}{10}\). \(\frac{11}{2}\)
A= \(\frac{11}{20}\)
Do 20> 19 => \(\frac{11}{20}\)< \(\frac{11}{19}\). Vậy A< \(\frac{11}{19}\)
Duyệt đi, chúc bạn học giỏi!
![](https://rs.olm.vn/images/avt/0.png?1311)
`A = 3/4 xx 8/9 xx ... xx 99/100`
`= (1xx3)/(2xx2) xx (2xx4)/(3xx3) xx ... xx (9xx11)/(10xx10)`
`= (1xx2xx3xx ... xx 9)/(2xx3xx...xx10) xx (3xx4xx5xx...xx 11)/(2xx3xx4xx...xx 10)`
`= 1/10 xx 11`
`= 11/10`.
Ta có: `11/10 > 1`
`11/19 < 1`.
`=> A > 11/19`.
![](https://rs.olm.vn/images/avt/0.png?1311)
đề bài bạn sai vì theo như quy luật thì :
A=\(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{81}+\dfrac{1}{100}\)
\(\dfrac{1}{4}>\dfrac{1}{3.2}\)
\(\dfrac{1}{9}>\dfrac{1}{3.4}\)
\(\dfrac{1}{16}>\dfrac{1}{4.5}\)
.
.
.
\(\dfrac{1}{81}>\dfrac{1}{9.10}\)
\(\dfrac{1}{100}>\dfrac{1}{10.11}\)
A > \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}+\dfrac{1}{10.11}\)
A > \(\dfrac{1}{2}+\dfrac{1}{11}\) =\(\dfrac{13}{22}\)
mà \(\dfrac{13}{22}\)>\(\dfrac{65}{132}\) ; A>\(\dfrac{13}{22}\)
Vậy A>\(\dfrac{65}{132}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = \(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{100}\)
= \(\frac{1}{4}+\left(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\right)\)
Ta có: \(\frac{1}{3^2}>\frac{1}{3.4}\)
\(\frac{1}{4^2}>\frac{1}{4.5}\)
.........
\(\frac{1}{10^2}>\frac{1}{10.11}\)
\(\Rightarrow A>\frac{1}{4}+\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}\right)\)
\(\Rightarrow A>\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(\Rightarrow A>\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{11}\right)=\frac{1}{4}+\frac{8}{33}=\frac{65}{132}\)
Vậy A > 65/132
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{81}\right)\left(1-\frac{1}{100}\right)\)
\(B=\frac{3}{4}\cdot\frac{8}{9}\cdot...\cdot\frac{80}{81}\cdot\frac{99}{100}\)
\(B=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot...\cdot\frac{8.10}{9.9}\cdot\frac{9.11}{10.10}\)
\(B=\frac{\left(1\cdot2\cdot...\cdot8\cdot9\right).\left(3\cdot4\cdot...\cdot10\cdot11\right)}{\left(2\cdot3\cdot..\cdot9\cdot10\right).\left(2\cdot3\cdot...\cdot9\cdot10\right)}\)
\(B=\frac{1\cdot2\cdot...\cdot8\cdot9}{2\cdot3\cdot...\cdot9\cdot10}\cdot\frac{3\cdot4\cdot...\cdot10\cdot11}{2\cdot3\cdot...\cdot9\cdot10}\)
\(B=\frac{1}{10}\cdot\frac{11}{2}=\frac{11}{20}\)
Vì 20 < 21 nên 11/20 > 11/21
Vậy .....
bạn vào link này nè:https://olm.vn/hoi-dap/question/980572.html
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)
\(\Leftrightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}+\frac{1}{10^2}\)
\(\Leftrightarrow A>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\)
\(\Leftrightarrow A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(\Leftrightarrow A>\frac{1}{2}-\frac{1}{11}\)
\(\Leftrightarrow A>\frac{9}{22}\)
Ta lại có:
\(\frac{9}{22}=\frac{9.11}{22\cdot11}=\frac{99}{132}\)
Ta thấy: 99>65
\(\Rightarrow\frac{99}{132}>\frac{65}{132}\)
\(\Rightarrow A>\frac{65}{132}\)
Vậy \(A>\frac{65}{132}\left(đpcm\right)\)
\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)
\(A=\frac{1}{4}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}+\frac{1}{10^2}\)
\(A>\frac{1}{4}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(A>\frac{1}{4}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
\(A>\frac{1}{4}+\frac{1}{3}-\frac{1}{11}\)
\(A>\frac{33}{132}+\frac{44}{132}-\frac{12}{132}\)
\(A>\frac{65}{132}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1/2*2+1/3*3+1/4*4+...+1/10*10.
A>1/1*2+1/2*3+1/3*4+...+1/9*10.
A>1-1/2+1/2-1/3+...+1/9-1/10.
A>1-1/10.
A>9/10.
=>A>1/2.
Mà 1/2=66/132>65/132.
=>A>65/132.
Vậy A>65/132.
A=1/2^2+1/3^2+1/4^2+......+1/9^2+1/10^2
=1/4+1/3×3+1/4×4+.....+1/9×9+1/10×10
=>A>1/4+(1/3×4+1/4×5+...+1/9×10+1/10×11)
=>A>1/4+(1/3-1/11)
=>A>1/4+8/33
=>A>65/132( đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\frac{3}{4}\cdot\frac{8}{9}\cdot\cdot\cdot\frac{99}{100}=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\cdot\cdot\frac{9\cdot11}{10\cdot10}=\frac{\left(1\cdot2\cdot\cdot\cdot\cdot9\right)\cdot\left(3\cdot4\cdot\cdot\cdot10\cdot11\right)}{\left(2\cdot3\cdot\cdot\cdot\cdot9\cdot10\right)\left(2\cdot3\cdot\cdot\cdot\cdot\cdot10\right)}=\frac{11}{2.10}=\frac{11}{20}\)\(