\(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+....+\frac{3}{418}+\frac{3}{550}\)

\(\Leftrightarrow\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{19.22}+\frac{3}{22.25}\)

\(\Leftrightarrow\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{19}-\frac{1}{22}+\frac{1}{22}-\frac{1}{25}\)

\(\Leftrightarrow\frac{1}{1}-\frac{1}{25}=\frac{24}{25}\)

Nhớ k cho m nhé!

\(3\times\left(\frac{x}{4}+\frac{x}{28}+\frac{x}{70}+\frac{x}{130}\right)=\frac{60}{13}\)

=> \(\frac{x}{4}+\frac{x}{28}+\frac{x}{70}+\frac{x}{130}=\frac{20}{13}\)

=> \(\frac{x}{1\cdot4}+\frac{x}{4\cdot7}+\frac{x}{7\cdot10}+\frac{x}{10\cdot13}=\frac{20}{13}\)

=> \(\frac{x}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}\right)=\frac{20}{13}\)

=> \(\frac{x}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{10}-\frac{1}{13}\right)=\frac{20}{13}\)

=> \(\frac{x}{3}\left(1-\frac{1}{13}\right)=\frac{20}{13}\)

=> \(\frac{x}{3}\cdot\frac{12}{13}=\frac{20}{13}\)

=> \(\frac{x}{3}=\frac{20}{13}:\frac{12}{13}=\frac{20}{13}\cdot\frac{13}{12}=\frac{5}{3}\)

=> x = 5

\(3\cdot\left(\frac{x}{4}+\frac{x}{28}+\frac{x}{70}+\frac{x}{130}\right)=\frac{60}{13}\)

\(3\cdot\left(\frac{x}{1\cdot4}+\frac{x}{4\cdot7}+\frac{x}{7\cdot10}+\frac{x}{10\cdot13}\right)=\frac{60}{13}\)

\(3\left(x-3\right)\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}\right)=\frac{60}{13}\)

\(\left(3x-9\right)\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}\right)=\frac{60}{13}\)

\(\left(3x-9\right)\left(1-\frac{1}{13}\right)=\frac{60}{13}\)

\(\left(3x-9\right)\cdot\frac{12}{13}=\frac{60}{13}\)

\(3x-9=\frac{\frac{60}{13}}{\frac{12}{13}}\)

\(3x-9=5\)

\(3x=5+9\)

\(3x=14\)

\(x=\frac{14}{3}\approx4,667\)

\(3.\left(\frac{x}{4}+\frac{x}{28}+\frac{x}{70}+\frac{x}{130}\right)=\frac{60}{13}\)

\(3.\left(\frac{x}{1.4}+\frac{x}{4.7}+\frac{x}{7.10}+\frac{x}{10.13}\right)=\frac{60}{13}\)

\(3.\frac{x}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}\right)=\frac{60}{13}\)

\(x.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}\right)=\frac{60}{13}\)

\(x.\left(1-\frac{1}{13}\right)=\frac{60}{13}\)

\(x.\frac{12}{13}=\frac{60}{13}\)

\(x=\frac{60}{13}\div\frac{12}{13}\)

\(x=5\)

\(B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)

\( B=\frac{1}{2}-\frac{1}{14}=\frac{3}{7}\)

\(C=\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+\frac{3}{208}=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\)

\(C=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\)

\(C=\frac{1}{1}-\frac{1}{16}=\frac{15}{16}\)

b) D = \(\frac{3}{4}+\frac{3}{8}+\frac{3}{70}+\frac{3}{130}+\frac{3}{208}+\frac{3}{304}\)

    D =  \(3\left(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{208}+\frac{1}{304}\right)\)

    D = \(3\left(\frac{1}{1x4}+\frac{1}{4x7}+\frac{1}{7x10}+\frac{1}{10x13}+\frac{1}{13x16}+\frac{1}{16x19}\right)\)

    D = \(\frac{1}{1}-\frac{1}{19}=\frac{18}{19}\)

Chắc vậy

A = \(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{56}\)

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{7.8}\)

  \(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{7}-\frac{1}{8}\)

  \(=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)

B = \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{11.13}\)

  \(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\)

  \(=1-\frac{1}{13}=\frac{12}{13}\)

\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{56}\)

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{7.8}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)

\(=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)

\(B=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{11.13}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\)

\(=1-\frac{1}{13}=\frac{12}{13}\)

A=\(\frac{1}{6}\)+\(\frac{1}{12}\)+......+\(\frac{1}{56}\)

A=\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{7.8}\)

A=\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+....+\(\frac{1}{7}\)-\(\frac{1}{8}\)

A=\(\frac{1}{2}\)-\(\frac{1}{8}\)=\(\frac{3}{8}\)

B=\(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+...+\(\frac{2}{11.13}\)

B=1-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+....+\(\frac{1}{11}\)-\(\frac{1}{13}\)

B=1-\(\frac{1}{13}\)=\(\frac{12}{13}\)

C=\(\frac{3}{4}\)+\(\frac{3}{28}\)+....+\(\frac{3}{304}\)

C=\(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+..+\(\frac{3}{16.19}\)

Rồi bạn cũng tách ra như câu A và câu B

Mọi người hướng dẫn mình làm bài này với 

\(S=\frac{3^2}{4}-\frac{3^2}{4.7}-\frac{3^2}{7.10}-...-\frac{3^2}{28.31}\)

\(S=\frac{3^2}{4}-\left(\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{28.31}\right)\)

\(S=\frac{9}{4}-3.\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{28.31}\right)\)

\(S=\frac{9}{4}-3.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{28}-\frac{1}{31}\right)\)

\(S=\frac{9}{4}-3.\left(1-\frac{1}{31}\right)\)

\(S=\frac{9}{4}-3.\frac{30}{31}=\frac{9}{4}-\frac{90}{31}=\frac{-81}{124}\)

Dấu bằng thứ 5 phải là vầy mới đúng chứ bạn ?
\(S=\frac{9}{4}-3.\left(\frac{1}{4}-\frac{1}{31}\right)\)

c)  \(A=\frac{6}{4}+\frac{6}{28}+\frac{6}{70}+\frac{6}{130}+\frac{6}{208}\) 

\(=\frac{6}{1.4}+\frac{6}{4.7}+\frac{6}{7.10}+\frac{6}{10.13}+\frac{6}{13.16}\) 

\(=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\)

\(=2\left(1-\frac{1}{16}\right)\) 

\(=2.\frac{15}{16}\) 

\(=\frac{15}{8}\) 

Vậy A=\(\frac{15}{8}\)

a) \(\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)

\(=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)

\(=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(=3\left(1-\frac{1}{100}\right)\)

\(=3.\frac{99}{100}=\frac{297}{100}\)

\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)

\(=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)

\(=\frac{1}{2}-\frac{1}{20}\)

\(=\frac{10}{20}-\frac{1}{20}\)

\(=\frac{9}{20}\)

a) \(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+\frac{3}{208}+\frac{3}{304}+\frac{3}{418}+\frac{3}{550}\)

= \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}+\frac{3}{19.22}+\frac{3}{22.25}\)

= \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+\frac{1}{19}-\frac{1}{22}+\frac{1}{22}-\frac{1}{25}\)

= \(\frac{1}{1}-\frac{1}{25}\)

= \(\frac{24}{25}\)

b) \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n+1\right).\left(2n+3\right)}\)

= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n+1}-\frac{1}{2n+3}\)

= \(\frac{1}{1}-\frac{1}{2n+3}\)

= \(\frac{2n+2}{2n+3}\)

c) \(\frac{7+\frac{7}{13}-\frac{7}{48}+\frac{7}{95}}{15+\frac{15}{13}-\frac{15}{48}+\frac{15}{95}}-\frac{7070707}{15151515}\)

= \(\frac{7\left(1+\frac{1}{13}-\frac{1}{48}+\frac{1}{95}\right)}{15\left(1+\frac{1}{13}-\frac{1}{48}+\frac{1}{95}\right)}-\frac{7.1010101}{15.1010101}\)

= \(\frac{7}{15}-\frac{7}{15}\)

= 0

a) 24/25

b) (2n+2)/(2n+3)

c) 0

sai thì thôi nhé

bài 1+2: phân tích mẫu thành nhân tử r` áp dụng 

1/ab=1/a-1/b 

bài 3+4: quy đồng rút gọn blah...