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B = 10/56 + 10/140 + 10/260 + ...+ 10/1400
B= 5/28 + 5/70 +.....+10/700
= 5/(4.7)+5/(7.10)+....5/(25.28)
3B= 5( 1/4 - 1/7 +1/7-1/10+......+1/25-1/28)
3B = 5 (1/4-1/28)
3B=15/14
B = 15/14 : 3
B = 5/14
![](https://rs.olm.vn/images/avt/0.png?1311)
S = 10/56 + 10/140 + 10/260 + ....... + 10/1400
S = 5/28 + 5/70 + 5/130 + 5/700
3S/5 = 3/4 x 7 + 3/7 x 10 + 30/10 x 13 + ....... + 3/25 x 28
3S/5 = 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13 + ........ + 1/25 - 1/28
3S/5 = 1/4 - 1/28
3S/5 = 3/14
S = 3/14 x 5/3
S = 5/14
Vậy S = 5/14
\(S=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+\frac{10}{1400}\)
\(S=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(S=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{25.28}\)
\(S=5.\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{25.28}\right)\)
\(S=5.\left(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{25.28}\right)\)
\(S=5.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(S=5.\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(S=5.\frac{3}{14}=\frac{15}{14}\)
Vậy \(S=\frac{15}{14}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
=\(\frac{5}{3}\times\left(\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+...+\frac{3}{700}\right)\)
=\(\frac{5}{3}\times\left(\frac{3}{4\times7}+\frac{3}{7\times10}+\frac{3}{10\times13}+...+\frac{3}{25\times28}\right)\)
=\(\frac{5}{3}\times\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{25}-\frac{1}{28}\right)\)
=\(\frac{5}{3}\times\left(\frac{1}{4}-\frac{1}{28}\right)\)
=\(\frac{5}{3}\times\frac{3}{14}\)
=\(\frac{5}{14}\)
Ko bít có đúng ko nhưng cứ thử nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
M = 10/56+10/140+10/260+10/1400
M= 5/28+5/70+5/130+5/700
3M/5=1/4-1/7+1/7-1/10+1/10-1/13+...+1/25
3M/5 = 3/14
M= 3/14+5/3=5/14
![](https://rs.olm.vn/images/avt/0.png?1311)
mk chỉ tính được kết qua thôi vì mk dụng máy tính casio =1931420 k mk nha
D = \(\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
D = \(\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(\frac{3D}{5}=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{25.28}\)
\(\frac{3D}{5}=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-...+\frac{1}{25}-\frac{1}{28}\)
\(\frac{3D}{5}=\frac{1}{4}-\frac{1}{28}\)
\(\frac{3D}{5}=\frac{3}{14}\)
\(D=\frac{3}{14}.\frac{5}{3}=\frac{5}{14}\)
Vậy tổng trên là : \(\frac{5}{14}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(=\frac{5}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{25.28}\right)\)
\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(=\frac{5}{3}\times\frac{3}{14}\)
\(=\frac{5}{14}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
S = 5/28 + 5/70 + 5/130 + ..... + 5/700
S = 5 / 4x7 + 5 / 7x10 + 5 / 10x13 + ..... + 5 / 25x28
SX3 / 5 = 3 / 4x7 + 3 / 7x10 + ....... + 3 / 25x28
Sx3/5=1/4-1/7+1/7-1/10+1/10+.....+1/25-1/28
Sx3/5=1/4+(1/7-1/7)+(1/9-1/9)+.....+(!/25-1/25)-1/28
Sx3/5=1/4-1/28
Sx3/5=3/14
S = 3/14: 3/5
S = 5/14
Vậy S = 5 /14
tk mk nhé !
\(C=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}=10\left(\frac{1}{56}+\frac{1}{140}+\frac{1}{260}+...+\frac{10}{1400}\right)\)
\(C=\frac{10}{2}\left(\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+...+\frac{1}{700}\right)=5\left(\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+...+\frac{1}{25\cdot28}\right)\)
\(C=5\cdot\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(C=\frac{5}{3}\cdot\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{3}\cdot\frac{3}{14}=\frac{5}{14}\)
\(B=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(B=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(B=\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{26.28}\right)\)
\(B=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{25}\right)\)
\(B=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(B=\frac{5}{3}.\frac{3}{14}\)
\(B=\frac{5}{14}\)