Đặt \(A=\frac{3}{5.7}+\frac{3}{7.9}+....+\frac{3}{59.61}\)

 \(\Rightarrow\frac{2}{3}.A=\frac{2}{3}.\left(\frac{3}{5.7}+\frac{3}{7.9}+....+\frac{3}{59.61}\right)\)

\(\Rightarrow\frac{2}{3}.A=\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{59.61}\)

\(\Rightarrow\frac{2}{3}.A=\frac{7-5}{5.7}+\frac{9-7}{7.9}+.....+\frac{61-59}{59.61}\)

\(\Rightarrow\frac{2}{3}.A=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{59}-\frac{1}{61}\)

\(\Rightarrow\frac{2}{3}.A=\frac{1}{5}-\frac{1}{61}\)

\(\Rightarrow\frac{2}{3}.A=\frac{56}{305}\)

\(\Rightarrow A=\frac{56}{305}:\frac{2}{3}\)

\(\Rightarrow A=\frac{56}{305}.\frac{3}{2}\)

\(\Rightarrow A=\frac{84}{305}\)

Vậy \(\frac{3}{5.7}+\frac{3}{7.9}+....+\frac{3}{59.61}=\frac{84}{305}\)

\(\frac{3}{5x7}+\frac{3}{7x9}+...+\frac{3}{59x61}\)

\(=\frac{3}{2}\left(\frac{2}{5x7}+\frac{2}{7x9}+...+\frac{2}{59x61}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}++...+\frac{1}{59}-\frac{1}{61}\right)\)

\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)\)

\(=\frac{3}{2}.\frac{56}{305}=\frac{84}{305}\)

Nguyễn Tuấn Minh giải đúng rồi nhé

\(\dfrac{6}{5.7}+\dfrac{6}{7.9}+...+\dfrac{6}{59.61}\)

\(=3\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\right)\)

\(=3\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)

\(=3\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)

\(=\dfrac{3.56}{305}\\ =\dfrac{168}{305}\)

\(=3\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)=3\cdot\left(\dfrac{1}{5}-\dfrac{1}{61}\right)=3\cdot\dfrac{56}{305}=\dfrac{168}{305}\)

Ta có :\(\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{59.61}=2.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)

\(=2\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)=2\left(\frac{1}{5}-\frac{1}{61}\right)=2.\frac{56}{305}=\frac{112}{305}\)

P/S : Dấu "." là dấu "x"

Bài làm:

Ta có: \(\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{59.61}\)

\(=2\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)

\(=2\left(\frac{1}{5}-\frac{1}{61}\right)\)

\(=2.\frac{56}{305}=\frac{112}{305}\)

\(\frac{4}{5\times7}+\frac{4}{7\times9}+.....+\frac{4}{59\times61}\)

\(=\frac{4}{2}\times\left(\frac{2}{5\times7}+\frac{2}{7\times9}+....+\frac{2}{59\times61}\right)\)

\(=2\times\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.....+\frac{1}{59}-\frac{1}{61}\right)\)

\(=2\times\left(\frac{1}{5}-\frac{1}{61}\right)\)

\(=2\times\frac{56}{305}\)

\(=\frac{112}{305}\)

Học tốt

bài này dễ mà 

C1 đặt 3 ra rồi nhân 2

C2 làm tắt nhân bằng phân số luôn thế thôi

\(S=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{2013.2015}\)

\(S=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2013.2015}\right)\)

\(S=\frac{3}{2}.\left(\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+...+\frac{2}{2013}-\frac{2}{2015}\right)\)

\(S=\frac{3}{2}.\left(\frac{2}{5}-\frac{2}{2015}\right)\)

\(S=\frac{3}{2}.\frac{804}{2015}\)

\(S=\frac{1206}{2015}\)

Giải:

\(B=\dfrac{3}{3\times5}+\dfrac{3}{5\times7}+\dfrac{3}{7\times9}+...+\dfrac{3}{48\times50}\) 

\(B=\dfrac{3}{2}\times\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{48\times50}\right)\) 

\(B=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\) 

\(B=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{50}\right)\)

\(B=\dfrac{3}{2}\times\dfrac{47}{150}\) 

\(B=\dfrac{47}{100}\) 

Chúc em học tốt!

917749738461936926399639748776398646491639394748947630373937366

\(\frac{3}{3\times5}+\frac{3}{5\times7}+\frac{3}{7\times9}+...+\frac{3}{99\times101}\)

\(=\frac{3}{2}\times\left(\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{99\times101}\right)\)

\(=\frac{3}{2}\times\left(\frac{5-3}{3\times5}+\frac{7-5}{5\times7}+\frac{9-7}{7\times9}+...+\frac{101-99}{99\times101}\right)\)

\(=\frac{3}{2}\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{3}{2}\times\left(\frac{1}{3}-\frac{1}{101}\right)\)

\(=\frac{49}{101}\)