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

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

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

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

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

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

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

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

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)

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

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 !!!! 

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

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

Ta có: \(\frac{1}{3^2}>\frac{1}{3.4}\)

\(\frac{1}{4^2}>\frac{1}{4.5}\)

...............

\(\frac{1}{10^2}< \frac{1}{10.11}\)

\(\Rightarrow A>\frac{1}{4}+\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}\right)=\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\right)=\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{100}\right)=\frac{1}{4}+\frac{8}{33}=\frac{65}{132}\)

Vậy A > 65/132

\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{10.10}\)

\(\Rightarrow A>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(A>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(A>1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{4}+\frac{1}{4}\right)+...+\left(-\frac{1}{9}+\frac{1}{9}\right)-\frac{1}{10}\)

\(A>1+0+0+0+...+0-\frac{1}{10}\)

\(A>1-\frac{1}{10}=\frac{9}{10}\)

\(\Rightarrow A>\frac{5}{10}=\frac{1}{2}\)

mà \(\frac{1}{2}=\frac{66}{132}>\frac{65}{132}\)

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

Vậy \(A>\frac{65}{132}\)

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

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

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

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

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

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

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

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

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

\(=\frac{1}{2^2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

\(=\frac{1}{2^2}+\frac{1}{3}-\frac{1}{11}\)

\(=\frac{65}{132}\)

vậy \(A>\frac{65}{132}\)

Ta có

A=122 +132 +142 +...+192 +1102 

A>122 +13.4 +14.5 +...+19.10 +110.11 

=122 +13 −14 +14 −15 +...+19 −110 +110 −111 

=122 +13 −111 

=65132 

vậy A>65132  

K CHO MK NHA 

To nham