Phân số thứ 3 you viết sai rồi

S₆ = 10/56 + 10/140 + 10/260 + ... + 10/1400

2/10.S₆ = 1/28 + 1/70 + 1/130 + ... + 1/700

6/10.S₆ = 3/4.7 + 3/7.10 + 3/10.13 + ... + 3/25.28

3/5.S₆ = 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13 + ... + 1/25 - 1/28

3/5.S₆ = 1/4 - 1/28 

3/5.S₆ = 7/28 - 1/28 = 3/14

Tự lm típ đc rồi nhé

\(S=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+.......+\frac{10}{1400}\)

\(S=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+\frac{5}{700}\)

\(\frac{3S}{5}=\frac{3}{4}\times7+\frac{3}{7}\times10+\frac{30}{10}\times13+........+\frac{3}{25}\times28\)

\(\frac{3S}{5}=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+......+\frac{1}{25}-\frac{1}{28}\)

\(\frac{3S}{5}=\frac{1}{4}-\frac{1}{28}\)

\(\frac{3S}{5}=\frac{3}{14}\)

\(S=\frac{3}{14}\times\frac{5}{3}\)

\(S=\frac{5}{14}\)

Vậy \(S=\frac{5}{14}\)

\(C=\frac{5}{4}+\frac{5}{28}+\frac{5}{70}+...+\frac{5}{297\cdot300}\)

\(C=5\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{297\cdot300}\right)\)

\(C=5\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{297}-\frac{1}{300}\right)\)

\(C=5\left(1-\frac{1}{300}\right)\)

\(C=5\cdot\frac{299}{300}\)

\(C=\frac{299}{60}\)

Doãn Hải Anh mk lớp 7 nên quên òi 

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)

\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\right)\)

\(2A=1-\frac{1}{3^{100}}\)

\(A=\frac{1-\frac{1}{3^{100}}}{2}\)

\(B=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)

\(B=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)

\(B=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{25.28}\)

\(3B=\frac{5.3}{4.7}+\frac{5.3}{7.10}+\frac{5.3}{10.13}+...+\frac{5.3}{25.28}\)

\(3B=5\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{25.28}\right)\)

\(3B=5\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)

\(3B=5\left(\frac{1}{4}-\frac{1}{28}\right)\)

\(3B=5\cdot\frac{3}{14}=\frac{15}{14}\)

\(B=\frac{15}{14}:3=\frac{5}{14}\)

a) \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)

\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\right)\)

\(2A=1-\frac{1}{3^{100}}\)

\(\Rightarrow A=\frac{1-\frac{1}{3^{100}}}{2}\)

b)  \(B=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)

\(B=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)

\(B=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{25.28}\)

\(B=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{5}{3}.\left(\frac{1}{7}-\frac{1}{10}\right)+\frac{5}{3}.\left(\frac{1}{10}-\frac{1}{13}\right)+...+\frac{5}{3}.\left(\frac{1}{25}-\frac{1}{28}\right)\)

\(B=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)

\(B=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)\)

\(B=\frac{5}{3}.\frac{3}{14}\)

\(\Rightarrow B=\frac{5}{14}\)

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)

\(3A-A=1-\frac{1}{3^{100}}\)\(\Rightarrow2A=\left(1-\frac{1}{3^{100}}\right)\Rightarrow A=\frac{1}{2}\times\left(1-\frac{1}{3^{100}}\right)\)

\(A=\dfrac{5}{28}+\dfrac{5}{70}+\dfrac{5}{130}+...+\dfrac{5}{700}\)

\(\dfrac{3A}{5}=\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{25.28}\)

\(\dfrac{3A}{5}=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\)

\(\dfrac{3A}{5}=\dfrac{1}{4}-\dfrac{1}{28}=\dfrac{3}{14}\)

⇒ \(A=\dfrac{5}{14}\)

bài này bạn chỉ cần tách mẫu ra là làm được thôi