\(\text{Ta có:}\) \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=\frac{2}{3}\)

\(\Leftrightarrow2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=\frac{2}{3}.2\)

\(\Leftrightarrow\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).x=\frac{4}{3}\)

\(\Leftrightarrow\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right).x=\frac{4}{3}\)

\(\Leftrightarrow\left(1-\frac{1}{11}\right)x=\frac{4}{3}\)

\(\Leftrightarrow\frac{10}{11}x=\frac{4}{3}\)

\(\Leftrightarrow x=\frac{4}{3}:\frac{10}{11}=\frac{22}{15}\)

Đặt A = \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{5}{31}\)

  2A   = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)

  2A   = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{\left(2x+1\right)}-\frac{1}{2x+3}=\frac{10}{31}\)

  2A   = \(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)

  Ta có :  \(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)

                           \(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)

                          \(\frac{1}{2x+3}=\frac{1}{93}\)

=> 2x + 3 = 93

     2x       = 90

       x       = 45

A = 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/2017. 2019

= ( 1 - 1/3 ) + ( 1/3 - 1/5 ) + ... + (1/2017 - 1/2019 )

= 1 - 1/2019

= 2018/2019

S = 1/31 + 1/32 +...+ 1/60

 Ta có các phân số : 1/31, 1/32, ..., 1/59 đều lớn hơn 1/60

 Nên S > 1/60 + 1/60 + 1/60 +...+ 1/60 ( có tất cả 30 phân số )

= 30/60 = 1/2

Vì 1/2 < 4/5 nên S <4/5

Vậy, chứng tỏ S < 4/5

Chúc bạn học tốt !

<=> 2/1.3 + 2/3.5 + 2/5.7 +....+ 2/(x+2)(x+4) = 100/101

<=> 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 +.....+ 1/x+2 - 1/x+4 = 100/101

<=> 1 - 1/x+4 = 100/101

<=> 1/x+4 = 1 - 100/101 <=> 1/x+4 = 1/101 <=> x+4 = 101 <=> x= 101 - 4 = 97

:)

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

~ Hok tốt ~

\(\)

Viết thành 2 . (1/3.5 + 1/5.7 + 1/7.9 + ...+ 1/97.99

Tui hk bít nữa

Mik giải phía dưới rồi đó. Câu lúc nãy bạn đăng ý

\(\left[\frac{12}{11}-\left(\frac{1}{2}+\frac{1}{44}\right)\right].\left(x-0,2\right)=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

\(\frac{25}{44}.\left(x-0,2\right)=\frac{1}{2}.\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{9.11}\right)\)

\(x-0,2=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right):\frac{25}{44}\)

\(x-\frac{1}{5}=\frac{22}{25}.\left(1-\frac{1}{11}\right)=\frac{22}{25}.\frac{10}{11}=\frac{4}{5}\)

\(x=\frac{4}{5}+\frac{1}{5}\)

\(x=1\)

\(\frac{1}{2}-\frac{1}{1.3}-\frac{1}{3.5}-\frac{1}{5.7}-\frac{1}{7.9}-\frac{1}{9.11}=\frac{4}{5}-x\)

<=> \(2.\frac{1}{2}-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)=\frac{8}{5}-2x\)

<=> \(1-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)=\frac{8}{5}-2x\)

<=> \(1-\left(1-\frac{1}{11}\right)-\frac{8}{5}=-2x\)

<=> \(-\frac{83}{55}=-2x\)

<=> \(x=\frac{83}{110}\)