\(B=\frac{5}{10.11}+\frac{5}{11.12}+\frac{5}{12.13}+...+\frac{5}{19.20}\)

\(B=5.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{19.20}\right)\)

\(B=5.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{19}-\frac{1}{20}\right)\)

\(B=5.\left(\frac{1}{10}-\frac{1}{20}\right)\)

\(B=5.\frac{1}{20}=\frac{1}{4}\)

\(C=\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\)

\(4C=4.\left(\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)

\(4C=\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}+\frac{4}{21.25}\)

\(4C=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\)

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

\(C=\frac{4}{25}:4=\frac{1}{25}\)

Câu B mình giải hơi bị lỗi xíu!! Mk viết lại nhé

B=5/10.11+5/11.12+5/12.13+...+5/19.20

B=5(1/10.11+1/11.12+1/12.13+...+1/19.20)

B=5(1/10-1/11+1/11-1/12+1/12-1/13+...+1/19-1/20)

B=5( 1/10-1/20)

B=5.1/20

B=1/4

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

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

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

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

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

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

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

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

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

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

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

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

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

`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`

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

\(A=8400\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)

\(=\frac{8400}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}+\frac{4}{21.25}\right)\)

\(=2100\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\right)\)

\(=2100\left(1-\frac{1}{25}\right)\)

\(=2100\cdot\frac{24}{25}\)

\(=2016\)

\(A=8400.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)

\(A=8400.\left(\frac{1.4}{1.5.4}+\frac{1.4}{5.9.4}+\frac{1.4}{9.13.4}+\frac{1.4}{13.17.4}+\frac{1.4}{17.21.4}+\frac{1.4}{21.25.4}\right)\)

\(A=8400.\frac{1}{4}.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)

\(A=8400.\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\right)\)

\(A=8400.\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{25}\right)\)

\(A=8400.\frac{1}{4}.\frac{24}{25}\)

\(A=2016\)

\(A=8400.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\right)\) 

\(A=8400.\left(1-\frac{1}{25}\right)\)

\(A=8400.\frac{24}{25}=8064\)

\(->A=8064\)

A= 5/10.11+5/11.12+5/12.13+5/13.14+5/14.15

A= (1/10.11+1/11.12+1/12.13+1/13.14+1/14.15) :5

A= [(1/10-1/11)+(1/11-1/12)+(1/12-1/13)+(1/13-1/14)+(1/14-1/15)] :5

A= (1/10-1/15):5

A= 1/30:5

A= 1/150

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

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

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

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

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

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

\(\Rightarrow x=2008\)

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

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

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

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

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

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

\(\Rightarrow x=15\)

bài này nâng cao lớp 6 mk giải rồi bạn nhờ ai giảng hộ nha nếu bn 5 lên 6

B=1/4.(4/1.5+4/5.9+......+4/25.29)

B=1/4.(1-1/5+1/5-1/9+.....+1/25-1/29)

B=1/4.(1-1/29)

B=1/4.28/29

B=7/29

\(B=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}+\frac{1}{25.29}\)

\(\Rightarrow4B=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}+\frac{4}{21.25}+\frac{4}{25.29}\)

\(\Rightarrow4B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}+\frac{1}{25}-\frac{1}{29}\)

\(\Rightarrow4B=1-\frac{1}{29}\)

\(\Rightarrow4B=\frac{29}{29}-\frac{1}{29}=\frac{28}{29}\)

\(\Rightarrow B=\frac{28}{29}:4=\frac{28}{29}.\frac{1}{4}=\frac{7}{29}\)

Vậy ....

\(B=\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+\frac{1}{17\cdot21}+\frac{1}{21\cdot25}+\frac{1}{25\cdot29}\)

\(B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{29}\)

\(B=1-\frac{1}{29}\)

\(B=\frac{28}{29}\)

Chúc bạn học tốt !