Ta có : 

                \(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}\)\(=1\)

                \(x+\left(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}\right)=1\)

               \(x+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)=1\)

               \(x+\left(\frac{1}{5}-\frac{1}{45}\right)=1\)

             \(x+\frac{8}{45}=1\)

             \(x=1-\frac{8}{45}=\frac{37}{45}\)

           Ủng hộ mk nha !!! ^_^

Chứng minh \(A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}< 1\)

\(A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\)

\(A=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\)

\(A=\frac{1}{1}-\frac{1}{21}\)

\(A=\frac{20}{21}\)

\(\frac{20}{21}< 1\)

=> \(A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}< 1\)( đpcm ) 

* Mình sợ sai xD *

a) A = 4/5.9 + 4/9.13 + 4/13.17 + ... + 4/41/45

A = 1/5 - 1/9 + 1/9 - 1/13 + 1/13 - 1/17 + ... + 1/41 - 1/45

A = 1/5 - 1/45

A = 8/45

b) B = ( 1 - 1/2 ) . ( 1 - 1/3 ) . ( 1 - 1/4 ) . ..... . ( 1 - 1/100 )

B = 1/2 . 2/3 . 3/4 . .... . 99/100

B = \(\frac{1.2.3.......99}{2.3.4......100}\)

B = 1/100

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

B = \(\frac{1}{2}.\frac{2}{3}.....\frac{99}{100}\)

B = \(\frac{1}{100}\)

\(\frac{x}{2008}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-...-\frac{1}{120}=\frac{5}{8}\)

\(\Rightarrow\frac{x}{2008}-[2\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)]=\frac{5}{8}\)

\(\Rightarrow\frac{x}{2008}-[2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)]=\frac{5}{8}\)

\(\Rightarrow\frac{x}{2008}-[2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{5}-\frac{1}{16}\right)]=\frac{5}{8}\)

\(\Rightarrow\frac{x}{2008}-[2.\frac{3}{16}]=\frac{5}{8}\)

\(\Rightarrow\frac{x}{2008}=1\)

\(\Rightarrow x=2008\)

\(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=\frac{29}{45}\)

\(\Rightarrow\frac{7}{x}+\left(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}\right)=\frac{29}{45}\)

\(\Rightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\Rightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\Rightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)

\(\Rightarrow\frac{7}{x}=\frac{21}{45}\)

\(\Rightarrow x=15\)

Ta có \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)(đk : \(x\ne0\))

=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)

=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)

=> \(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)

=> \(\frac{7}{x}=\frac{7}{15}\)

=> x = 15 (tm)

b) \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)

=> \(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}\right)=\frac{15}{93}\)

=> \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)

=> \(\frac{1}{3}-\frac{1}{n+3}=\frac{10}{31}\)

=> \(\frac{1}{2x+3}=\frac{1}{93}\)

=> 2x + 3 = 93

=> 2x = 90

=> x = 45 

= 4/2.4 + 4/4.6 + 4/6.8 + ... + 4/99.100

= 2/2 - 2/4 + 2/4 - 2/6 + 2/6 - 2/8 + ... + 2/99 - 2/100

= 2/2 - 2/100

= 98/100 = 49/50

\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{41.45}\)

\(=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)\)

\(=\frac{1}{4}.\left(1-\frac{1}{45}\right)\)

\(=\frac{1}{4}.\frac{44}{45}\)

\(=\frac{11}{45}\)

Đặt \(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\)  là A.

Ta có:

\(A=\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\)

\(4A=4\left(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\right)\)

\(4A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}\)

\(4A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\)

\(4A=1-\frac{1}{45}\)

\(4A=\frac{44}{45}\)

\(A=\frac{44}{45}:4\)

\(A=\frac{11}{45}\)

Vậy \(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}=\frac{11}{45}\)

cảm ơn 2 bn nhé

mk kết bạn nha