A= 1/5.9+1/9.13+1/13.17+1/17.21+1/21.25

4A= 4/5.9+4/9.13+4/13.17+4/17.21+4/21.25

4A= (1/5-1/9)+(1/9-1/13)+(1/13-1/17)+(1/17-1/21)+(1/21-1/25)

4A= 1/5- 1/25

4A= 4/25

A= 4/25 :4 

A= 1/25

\(4S=4.\left(\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{21.25}\right)\)

=\(\frac{4}{5.9}+\frac{4}{9.13}+....+\frac{4}{21.25}_{ }\)

=\(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+....+\frac{1}{21}-\frac{1}{23}\)

=\(\frac{1}{5}-\frac{1}{25}=\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)

=> \(S=\frac{4}{25}:4=\frac{4}{25}.\frac{1}{4}=\frac{1}{25}\)

\(S=\frac{1}{5\times9}+\frac{1}{9\times13}+...+\frac{1}{21\times25}\)

\(S\times4=\frac{4}{5\times9}=\frac{4}{9\times13}+...+\frac{4}{21\times25}\)

\(S\times4=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{21}-\frac{1}{25}\)

\(S\times4=\frac{1}{5}-\frac{1}{25}\)

\(S\times4=\frac{4}{25}\)

\(S=\frac{1}{25}\)

ta có S=  1/5.9+1/9.13+1/13.17+1/17.21+1/21.25

<=>4S=4.(1/5.9+1/9.13+1/13.17+1/17.21+1/21.25)

<=>4S=4/5.9+4/9.13+4/13.17+4/17.21+4/21.25

<=>4S=1/5-1/9+1/9-1/13+1/13-1/17+1/21-1/25

<=>4S=1/5-1/25

<=>4S=4/25

<=>S=4/25:4

<=>S=1/25

vậy S=1/25

Tính chất của phân số bạn cần biết như sau:

\(\dfrac{b-a}{a\cdot b}=\dfrac{1}{a}-\dfrac{1}{b}\)

Gọi biểu thức trên là A ,ta có:

\(A=\dfrac{1}{5\cdot9}+\dfrac{1}{9\cdot13}+\dfrac{1}{13\cdot17}+\dfrac{1}{17\cdot21}+\dfrac{1}{21\cdot25}\)

\(4A=\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+\dfrac{4}{13\cdot17}+\dfrac{4}{17\cdot21}+\dfrac{4}{21\cdot25}\)

\(4A=\dfrac{9-5}{5\cdot9}+\dfrac{13-9}{9-13}+\dfrac{17-13}{13\cdot17}+\dfrac{21-17}{17\cdot21}+\dfrac{25-21}{21\cdot25}\)

Áp dụng tính chất phân số đã nêu ở trên, ta được:

\(4A=\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{25}\)

\(4A=\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{5}{25}-\dfrac{1}{25}=\dfrac{4}{25}\)

\(A=4A:4=\dfrac{4}{25}:4=\dfrac{16}{25}\)

Vậy \(A=\dfrac{16}{25}\)

\(\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{21.25}\\ =\dfrac{4\cdot\dfrac{1}{4}}{5.9}+\dfrac{4\cdot\dfrac{1}{4}}{9.13}+...+\dfrac{4\cdot\dfrac{1}{4}}{21.25}\\ =\dfrac{1}{4}\left(\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{21.25}\right)\\ =\dfrac{1}{4}\cdot\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{21}-\dfrac{1}{25}\right)\\ =\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{25}\right)=\dfrac{1}{4}\left(\dfrac{5}{25}-\dfrac{1}{25}\right)\\ =\dfrac{1}{4}\cdot\dfrac{4}{25}=\dfrac{1}{25}\)

`1/(5.9) + 1/(9.13) + ...+ 1/(21.25)`

`= 1/5 - 1/9 + 1/9 - 1/13 + ... + 1/21 - 1/25`

`= 1/5 - 1/25`

`= 4/25`

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

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

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

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

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

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

                      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 !!! ^_^

a)   \(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+\frac{1}{8\cdot10}\)

\(=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)

\(=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(=\frac{1}{2}\cdot\frac{2}{5}=\frac{2}{10}=\frac{1}{5}\)

b)   \(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}\)

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

\(=1-\frac{1}{17}=\frac{16}{17}\)

hok tốt ...

a)\(A=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+\frac{1}{8\cdot10}\)

\(2A=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+\frac{2}{8\cdot10}\)

\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)

\(A=\frac{2}{5}\cdot\frac{1}{2}=\frac{1}{5}\)

b)\(B=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}=1-\frac{1}{17}=\frac{16}{17}\)

a) 1/2.4 + 1/4.6 + 1/6.8 + 1/8.10

= 2/2.2.4 + 2/2.4.6 + 2/2.6.8 + 2/2.8.10 ( nhân cả tử và mẫu với 2)

= 1/2 .( 2/2.4 + 2/4.6 + 2/6.8 + 2/8.10 )

= 1/2 .(1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/10)

= 1/2.(1/2 - 1/10)

= 1/2.( 5/10 - 1/10) = 1/2.4/10 = 2/10 = 1/5

b) 4/1.5+ 4/5.9 + 4/ 9.13 + 4/13.17

= 1- 1/5 + 1/5 - 1/9 + 1/9 - 1/13 + 1/13 - 1/17

= 1- 1/17

= 16/17 

\(S1=\dfrac{5}{10.11}+\dfrac{5}{11.12}+.............+\dfrac{5}{14.15}\)

\(\Leftrightarrow S1=5\left(\dfrac{1}{10.11}+\dfrac{1}{11.12}+...............+\dfrac{1}{14.15}\right)\)

\(\Leftrightarrow S1=5\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+.............+\dfrac{1}{14}-\dfrac{1}{15}\right)\)

\(\Leftrightarrow S1=5\left(\dfrac{1}{10}-\dfrac{1}{15}\right)\)

\(\Leftrightarrow S1=5.\dfrac{1}{30}=\dfrac{1}{6}\)

\(S2=\dfrac{1}{5.9}+\dfrac{1}{9.13}+\dfrac{1}{13.17}+........+\dfrac{1}{21.25}\)

\(\Leftrightarrow4S_2=\dfrac{4}{5.9}+\dfrac{4}{9.13}+..............+\dfrac{4}{21.25}\)

\(\Leftrightarrow4S_2=\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+............+\dfrac{1}{21}-\dfrac{1}{25}\)

\(\Leftrightarrow4S_2=\dfrac{1}{5}-\dfrac{1}{25}\)

\(\Leftrightarrow4S_2=\dfrac{4}{25}\)

\(\Leftrightarrow S_2=\dfrac{16}{25}\)

Ta có : \(\left(x-1\right)^2+\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{41.45}=\dfrac{49}{900}\)

\(\Leftrightarrow\left(x-1\right)^2+\dfrac{1}{4}.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\right)=\dfrac{49}{900}\)

\(\Leftrightarrow\left(x-1\right)^2+\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{45}\right)=\dfrac{49}{900}\)

\(\Leftrightarrow\left(x-1\right)^2=\dfrac{1}{100}\)  \(\Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{10}\\x-1=-\dfrac{1}{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)

Vậy ...