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21 tháng 3 2019

\(A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)....\left(1-\frac{1}{100}\right)=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.......\frac{99}{100}=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}.....\frac{9.11}{10^2}=\frac{\left(1.2.3....9\right).\left(3.4.5....11\right)}{\left(2.3.4....10\right).\left(2.3.4....10\right)}=\frac{1.11}{10.2}=\frac{11}{20}\)

21 tháng 3 2019

Cảm ơn bạn nhiều yeu

14 tháng 5 2018

Ta có :

 \(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)

\(=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}+\frac{1}{10^2}\)

\(\Rightarrow A>\frac{1}{2^2}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}\)

\(\Rightarrow A>\frac{1}{4}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

\(\Rightarrow A>\frac{1}{4}+\frac{1}{3}-\frac{1}{11}\)

\(\Rightarrow A>\frac{65}{132}\left(đpcm\right)\)

Chúc bạn học tốt !!!! 

8 tháng 5 2017

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 > \(\dfrac{1}{2}+\dfrac{1}{11}\) =\(\dfrac{13}{22}\)

\(\dfrac{13}{22}\)>\(\dfrac{65}{132}\) ; A>\(\dfrac{13}{22}\)

Vậy A>\(\dfrac{65}{132}\)

6 tháng 5 2018

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 \(A>\frac{65}{132}\left(đpcm\right)\)

6 tháng 5 2018

\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)

\(A=\frac{1}{4}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}+\frac{1}{10^2}\)

\(A>\frac{1}{4}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}\)

\(A>\frac{1}{4}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)

\(A>\frac{1}{4}+\frac{1}{3}-\frac{1}{11}\)

\(A>\frac{33}{132}+\frac{44}{132}-\frac{12}{132}\)

\(A>\frac{65}{132}\)

4 tháng 5 2017

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.

6 tháng 5 2017

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)

4 tháng 5 2017

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