Ta có : A = 1 / (1.2) + 1 / (3.4) + ... + 1 / (99.100) > 1 / (1.2) + 1 / (3.4) = 1 / 2 + 1 / 12 = 7 / 12 (1)
Lại có : A = 1 / (1.2) + 1 / (3.4) + ... + 1 / (99.100) = (1 - 1 / 2) + (1 / 3 - 1 / 4) + ... + (1 / 99 - 100)

                =  (1 - 1 / 2 + 1 / 3) - (1 / 4 - 1 / 5) - (1 / 6 - 1 / 7) - ... - (1 / 98 - 1 / 99) - 1 / 100 <  1 - 1 / 2 + 1 / 3 = 5 / 6 (2)
Từ (1) và (2) => 7 / 12 < A < 5 / 6

@Lê Nguyên Hạo

nhận dc thông báo ko

A = 1/1x2 + 1/3x4 + 1/4x5 + 1/5x6 + ..... + 1/99x100

A = 1 - 1/2 + 1/2 -1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + .... + 1/99 - 1/100

A = 1 - 1/100

A = 99/100

Truong Quang Minh vào đây tham khảo nha:/hoi-dap/question/119017.html

A = 1/1x2 + 1/3x4 + 1/4x5 + 1/5x6 + ..... + 1/99x100
A = 1 - 1/2 + 1/2 -1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + .... + 1/99 - 1/100
A = 1 - 1/100
A = 99/100
 

1

Ta có :A=1/1.2+1/3.4+...+1/99.100=1/2+1/12+...+1/9900

           7/12=1/2+1/12

Vì 1/2+1/12<1/2+1/12+...+1/9900

Nên: 7/12<A (1)

Lại có:A=1/1.2+1/3.4+...+1/99.100

              =1-1/2+1/3-1/4+...+1/99-1/100

              =(1-1/2+1/3)+(-1/4+1/5-1/6)+...+(-1/98+1/99-1/100)

          5/6=1-1/2+1/3

vì: 1-1/2+1/3  <  (1-1/2+1/3)+(-1/4+1/5-1/6)+...+(-1/98+1/99-1/100)

nên    5/6        <  A     (2)

Từ (1) và (2) suy ra 7/12<A<5/6

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

Ta có : \(\frac{1}{51}>\frac{1}{52}>...>\frac{1}{75},\frac{1}{76}>\frac{1}{77}>...>\frac{1}{100}\)nên :

\(A>\frac{1}{75}.25+\frac{1}{100}.25=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

\(A< \frac{1}{51}.25+\frac{1}{76}.25< \frac{1}{50}.25+\frac{1}{75}.25=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)

Vậy \(\frac{7}{12}< A< \frac{5}{6}\)

\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=1+\frac{1}{2}+...+\frac{1}{100}-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

\(=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}\right)\)

Ta có: \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}>\frac{1}{75}+\frac{1}{75}+...+\frac{1}{75}=\frac{25}{75}=\frac{1}{3}\)

\(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{25}{100}=\frac{1}{4}\)

\(\Rightarrow A>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\left(1\right)\)

Lại có: \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}< \frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}=\frac{25}{50}=\frac{1}{2}\)

\(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}< \frac{1}{75}+\frac{1}{75}+...+\frac{1}{75}=\frac{25}{75}=\frac{1}{3}\)

\(\Rightarrow A< \frac{1}{2}+\frac{1}{3}=\frac{5}{6}\left(2\right)\)

Từ (1) và (2) => \(\frac{7}{12}< A< \frac{5}{6}\left(đpcm\right)\)

\(A=\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{49.50}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=\frac{1}{1}-\frac{1}{50}\)

\(A=\frac{49}{50}\)

Vì \(\frac{245}{420}< \frac{245}{294}< \frac{245}{250}\)

Vậy \(\frac{7}{12}< \frac{49}{50}< \frac{5}{6}\)

\(A=\frac{1}{2}+\frac{1}{12}+...+\frac{1}{9900}>\frac{1}{2}+\frac{1}{12}=\frac{7}{12}\)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\left(1-\frac{1}{2}+\frac{1}{3}\right)-\left(\frac{1}{4}-\frac{1}{5}\right)-...-\left(\frac{1}{98}-\frac{1}{99}\right)-\frac{1}{100}<\left(1-\frac{1}{2}+\frac{1}{3}\right)=\frac{5}{6}\)

Vậy đpcm