Ta có : 

\(A=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{9.9}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\)

\(=1-\frac{1}{9}=\frac{8}{9}\Rightarrow A< \frac{8}{9}\)(1)

Lại có \(A=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{9.9}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9.10}=\frac{1}{2}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\Rightarrow A>\frac{2}{5}\)(2)

Từ (1) (2) => \(\frac{2}{5}< A< \frac{8}{9}\left(\text{ĐPCM}\right)\)

                         Bài làm :

Ta có :

\(A=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{9.9}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

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

\(A>\frac{1}{2}-\frac{1}{10}\)

\(A>\frac{2}{5}\left(1\right)\)

Ta cũng có  : 

\( A=\frac{1}{2.2}+\frac{1}{3.3}+......+\frac{1}{9.9}< \frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{8.9}\)

\(A< 1-\frac{1}{2}+\frac{1}{2}-......+\frac{1}{8}-\frac{1}{9}\)

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

\(A< \frac{8}{9}\left(2\right)\)

\(\text{Từ (1) và (2) }\Rightarrow\frac{2}{5}< A< \frac{8}{9}\)

=> Điều phải chứng minh

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

A= 1/2.2 + 1/3.3 + 1/4.4 + 1/5.5 + 1/6.6 + 1/7.7 + 1/8.8 + 1/9.9

Vì 1/2.2 > 1/2.3; 1/3.3 > 1/3.4 ; 1/5.5 > 1/5.6;...... nên 

1/2.2 +1/3.3 + 1/4.4 + 1/5.5 + 1/6.6 + 1/7.7 + 1/8.8 + 1/9.9 > 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10

Ta có: 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10

= 1/2-1/3 + 1/3 -1/4 + 1/4-1/5+...+1/9-1/10

= 1/2- 1/10

= 2/5

Vì A < 2/5 mà 2/5 <7/8 nên 2/5 < A < 7/8

Vậy....

Bài làm:

Ta có: \(S=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}\)

\(>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{9.10}\)

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

\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)\(\Rightarrow\frac{2}{5}< S\)

Cái còn lại tự CM

\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)

\(2A=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\)

\(2A+A=\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\right)+\left(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\right)\)

\(3A=1-\frac{1}{64}\)

\(3A=\frac{63}{64}\Rightarrow A=\frac{63}{64}\div3=\frac{21}{64}< \frac{1}{3}\)

bài 6 \(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+\ldots+\frac{2}{80\times82}+\frac{2}{82\times84}\)

= \(\frac12-\frac14+\frac14-\frac16+\frac16-\cdots+\frac{1}{80}-\frac{1}{82}+\frac{1}{82}-\frac{1}{84}\)

= \(\frac12-\left(\frac14-\frac14\right)-\left(\frac16-\frac16\right)-\cdots-\left(\frac{1}{82}-\frac{1}{82}\right)-\frac{1}{84}\)

=\(\frac12-0-0-\frac{1}{84}\)

=\(\frac12\) - \(\frac{1}{84}\)

=\(\frac{42}{84}-\frac{1}{84}\)

=\(\frac{41}{84}\)

Bài 7

a)\(\frac{19}{4}=4\frac34\)

b)\(\frac{27}{5}=5\frac25\)

c)\(\frac{56}{8}=7\frac08\)