xí nha!đây ko phải toán lớp 5 nha bạn !

cô giáo mình ra đề như vậy mà

\(=\frac{1}{1.3}.\frac{1}{2.4}...\frac{1}{9.11}=\frac{1}{1.2.3^2...9^2.10.11}\)

bài này lớp 6 tui gặp nè ^-^

Làm lại đề cho:

\(\left(\frac{1}{4}-1\right)\cdot\left(\frac{1}{9}-1\right)\cdot\left(\frac{1}{16}-1\right)...\left(\frac{1}{81}-1\right)\cdot\left(\frac{1}{100}-1\right)\)

Tính nhẩm

TOI KO BIET

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

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

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

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

\(\Leftrightarrow A>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\)

\(\Leftrightarrow A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

\(\Leftrightarrow A>\frac{1}{2}-\frac{1}{11}\)

\(\Leftrightarrow A>\frac{9}{22}\)

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

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

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

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

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

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

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

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