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

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

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

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

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

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

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

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

sao dễ vậy

mình cũng cần làm bài này!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!\(HELPME\)

ta có:

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

mk chỉ lm đến đó thui! mk còn lm nhưng mk lm chó nhanh thì bn tự lm nha!

Ta có :

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

\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{92}+\frac{1}{10^2}\)

Mà \(\frac{1}{3^2}>\frac{1}{3.4}\)

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

\(...\)

\(\frac{1}{9^2}>\frac{1}{9.10}\)

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

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

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

\(\Rightarrow A-\frac{1}{2^2}>\frac{1}{3}-\frac{1}{11}\)

\(\Rightarrow A-\frac{1}{4}>\frac{8}{33}\)

\(\Rightarrow A>\frac{8}{33}+\frac{1}{4}\)

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

\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}+\frac{1}{10.10}\)

\(A>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.10}\)

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

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

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

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

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

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

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

Vậy \(A>\frac{65}{132}\)

phải là A = biểu thức đó và A = 9/10

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

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

A= 0

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

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

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

\(A>\frac{1}{2.2}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}+\frac{1}{10.11}\)

\(=\frac{1}{2.2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+\frac{1}{5}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)

\(=\frac{1}{2.2}+\frac{1}{3}-\frac{1}{11}\)

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

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

tất

nhiên

là lm

đc 

nhìn đã biết đc quy ;uật r ko cần phải đọc lâu lm j

mình định lm y hệt như lê mạnh tiến đạt

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

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

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

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

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

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

\(A=\left(\frac{1-2^2}{2^2}\right)\left(\frac{1-3^2}{3^2}\right)....\left(\frac{1-10^2}{10^2}\right)\)

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

\(A=\frac{3.\left(-1\right)}{2^2}.\frac{\left(-2\right).4}{3^2}........\frac{\left(-9\right).11}{10^2}=-\left(\frac{1.3}{2^2}.\frac{2.4}{3^2}....\frac{9.11}{10^2}\right)\)

\(=-\left(\frac{1.2....9}{2.3....10}.\frac{3.4....11}{2.3.4...10}\right)=-\left(\frac{1}{10}.\frac{11}{2}\right)=\frac{-11}{20}\)

a, Gọi biểu thức đó là A

Ta có :

A = \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)

A x 3 = \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}-\frac{1}{729}\)

A x 3 = \(1+A-\frac{1}{729}\)

A x 3 = \(\frac{728}{729}+A\)

A x 2 + A = \(\frac{728}{729}+A\)

A x 2 = \(\frac{728}{729}\)(bỏ A ở cả 2 vế)

A = \(\frac{728}{729}\div2=\frac{364}{729}\)

Đáp án = \(\frac{364}{729}\)

b, Phần này mình nghĩ là bạn sai đề rồi. Phải là \(\frac{45\times16-17}{45\times15+28}\)