\(B=2\left(\frac{5}{7.12}+\frac{5}{12.17}+...+\frac{5}{502.507}\right)\)

\(B=2\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{502}-\frac{1}{507}\right)\)

\(B=2\left(\frac{1}{7}-\frac{1}{507}\right)\)

\(B=2\times\frac{500}{3549}\)

\(B=\frac{1000}{3549}\)

\(B=\frac{10}{7.12}+\frac{10}{12.17}+\frac{10}{17.22}+......+\frac{10}{502.507}\)

\(B=\frac{10}{5}.\left(\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{17.22}+........+\frac{5}{502.507}\right)\)

\(B=\frac{10}{5}.\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}+.....+\frac{1}{502}-\frac{1}{507}\right)\)

\(B=\frac{10}{5}.\left(\frac{1}{7}-\frac{1}{507}\right)=2.\frac{500}{3549}=\frac{1000}{3549}\)

\(K=\frac{10}{7.12}+\frac{10}{12.17}+\frac{10}{17.22}+...+\frac{10}{502.507}\)

\(\Leftrightarrow K=\frac{10}{5}\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}+...+\frac{1}{502}-\frac{1}{507}\right)\)

\(\Leftrightarrow K=2\left(\frac{1}{7}-\frac{1}{507}\right)\)

\(\Leftrightarrow K=2\cdot\frac{500}{3549}\)

\(\Leftrightarrow K=\frac{1000}{3549}\)

5xK=5/7.12+5/12.17+..............+5/502.507

5xK=(1/7-1/12)+(1/12-1/17)+........+(1/502-1/507)

5xK=1/7-1/507

5xK=500/3549

K   =500/3549:5

K    =100/3549

\(N=2015+\frac{10}{2.7}+\frac{10}{7.12}+\frac{10}{12.17}+\frac{10}{17.22}\)

     \(=2\left(1007,5+\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{17.22}\right)\)

     \(=2\left(1007,5+\frac{7-2}{2.7}+\frac{12-7}{7.12}+\frac{17-12}{12.17}+\frac{22-17}{17.22}\right)\)

     \(=2\left(1007,5+\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}\right)\)

     \(=2\left(1007,5+\frac{1}{2}-\frac{1}{22}\right)\)

     \(=2015+1-\frac{1}{11}\)

     \(=\frac{22175}{11}\)

N = \(2015+\frac{10}{2,7}+\frac{10}{7,12}+\frac{10}{12,17}+\frac{10}{17,22}=2021.510611\)

\(N=2015+\frac{10}{2\cdot7}+\frac{10}{7\cdot12}+\frac{10}{12\cdot17}+\frac{10}{17\cdot22}=2015+2\left(\frac{5}{2\cdot7}+\frac{5}{7\cdot12}+\frac{5}{12\cdot17}+\frac{5}{17\cdot22}\right)\)\(=2015+2\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}\right)\)

\(=2015+2\left[\left(\frac{1}{2}-\frac{1}{22}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{17}-\frac{1}{17}\right)\right]\)

\(=2015+2\left[\left(\frac{11}{22}-\frac{1}{22}\right)+0+...+0\right]=2015+2\cdot\frac{10}{22}=2015+\frac{10}{11}\)

xong rồi đó, cái còn lại thì bạn quy đồng rồi cộng nha

mk cũng ko chắc do tính nhẩm

\(\Rightarrow C=\frac{10}{5}\left(\frac{1}{7.12}+\frac{1}{12.17}+\frac{1}{17.22}+...+\frac{1}{502.507}\right)\)

\(\Rightarrow C=2.\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+....+\frac{1}{507}-\frac{1}{507}\right)\)

\(\Rightarrow C=2.\left(\frac{1}{7}-\frac{1}{507}\right)=2.\frac{1}{7}-2.\frac{1}{507}=\frac{2}{7}-\frac{2}{507}\)

\(A=\frac{1}{7\cdot12}+\frac{1}{12\cdot17}+\frac{1}{17\cdot22}+...+\frac{1}{52\cdot57}\)

\(A=\frac{1}{5}\left(\frac{5}{7\cdot12}+\frac{5}{12\cdot17}+\frac{5}{17\cdot22}+...+\frac{5}{52\cdot57}\right)\)

\(A=\frac{1}{5}\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{52}-\frac{1}{57}\right)\)

\(A=\frac{1}{5}\left(\frac{1}{7}-\frac{1}{57}\right)=\frac{1}{5}\cdot\frac{50}{399}=\frac{10}{399}\)

\(B=\frac{10}{8\cdot13}+\frac{10}{13\cdot18}+\frac{10}{18\cdot23}+...+\frac{10}{253\cdot258}\)

\(B=\frac{10}{5}\left(\frac{5}{8\cdot13}+\frac{5}{13\cdot18}+\frac{5}{18\cdot23}+...+\frac{5}{253\cdot258}\right)\)

\(B=2\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{253}-\frac{1}{258}\right)\)

\(B=2\left(\frac{1}{8}-\frac{1}{258}\right)=2\cdot\frac{125}{1032}=\frac{125}{516}\)

*Cái đây giải thích hơi bị " khó hiểu " :

Chỗ mẫu (12 - 7) = (17 - 12) = ... = (57 - 52) = 5

Tử là 1 , mẫu là 5 nên tử/mẫu = 1/5

Hay \(\frac{1}{5}\left(\frac{5}{7\cdot12}+\frac{5}{12\cdot17}+...+\frac{5}{52\cdot57}\right)\)

Còn bạn Trương Bùi Linh thì :

Mẫu = (13 - 8) = (18 - 13) = (23 - 18) = ... = 5

Tử là 10,mẫu là 5 => tử / mẫu = 10/5 = 2

D = \(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{2006.2009}\)

= \(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{2006}-\dfrac{1}{2009}\)

= \(\dfrac{1}{5}-\dfrac{1}{9}=\dfrac{2004}{10045}\)

C = \(\dfrac{10}{7.12}+\dfrac{10}{12.17}+\dfrac{10}{17.22}+...+\dfrac{10}{502.507}\)

= \(\dfrac{10}{5}\left(\dfrac{5}{7.12}+\dfrac{5}{12.17}+\dfrac{5}{17.22}+...+\dfrac{5}{502.507}\right)\)

= \(\dfrac{10}{5}\left(\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{22}+....+\dfrac{1}{502}-\dfrac{1}{507}\right)\)

= \(\dfrac{10}{5}\left(\dfrac{1}{5}-\dfrac{1}{507}\right)\)

= \(\dfrac{10}{5}.\dfrac{502}{2535}\)

= \(\dfrac{1000}{3549}\)

C=1000/3549
D=2004/10045

\(\left(\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{17.22}+\frac{5}{22.27}+\frac{5}{27.32}\right)x=945\)

\(\left(\frac{7-2}{2.7}+\frac{12-7}{7.12}+\frac{17-12}{12.17}+\frac{22-17}{17.22}+\frac{27-22}{22.27}+\frac{32-27}{27.32}\right)x=945\)

\(\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{27}-\frac{1}{32}\right)x=945\)

\(\left(\frac{1}{2}-\frac{1}{32}\right)x=945\Leftrightarrow\frac{15}{32}x=945\Leftrightarrow x=945:\frac{15}{32}=2016\)

Sửa: 

\(B=\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{17.22}+\frac{5}{22.27}\)

Trả lời

\(B=\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12\cdot17}+\frac{5}{17\cdot22}+\frac{5}{22\cdot27}\)

\(\Rightarrow B=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}+\frac{1}{22}-\frac{1}{27}\)

\(\Rightarrow B=\frac{1}{2}-\frac{1}{27}\)

\(\Rightarrow B=\frac{25}{54}\)

Vậy B=\(\frac{25}{54}\)