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

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

\(S=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.60}\right)\)

\(S=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{60}\right)\)

\(S=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{60}\right)\)

\(S=\frac{3}{2}.\left(\frac{12}{60}-\frac{1}{60}\right)\)

\(S=\frac{3}{2}.\frac{11}{60}\)

\(S=\frac{11}{40}\)

nhi phương k cho mk đi

\(S=\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+...+\frac{3}{59\cdot61}\)

\(S=\frac{3}{2}\left[\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{59\cdot61}\right]\)

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

\(S=\frac{3}{2}\left[\frac{1}{5}-\frac{1}{61}\right]=\frac{3}{2}\cdot\frac{56}{305}=\frac{84}{305}\)

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

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

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

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

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

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

18/65

18/65 nhé  bạn

nêu cách giải ra mình mới hiểu dc

\(\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\right).y=\frac{2}{3}\)

\(\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)y=\frac{2}{3}\)

\(\left(1-\frac{1}{11}\right).y=\frac{2}{3}\)

\(\frac{10}{11}.y=\frac{2}{3}\)

\(y=\frac{2}{3}.\frac{11}{10}\)

\(y=\frac{22}{30}\)

\(\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\right).y=\frac{2}{3}\)

         \(\frac{10}{11}.y=\frac{2}{3}\)

                    \(y=\frac{11}{15}\)

\(\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).y=\frac{2}{3}\)

\(\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{11-9}{9.11}\right).y=\frac{2}{3}\)

\(\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)

\(\left(\frac{1}{1}-\frac{1}{11}\right).y=\frac{2}{3}\)

\(\frac{10}{11}.y=\frac{2}{3}\)

         \(y=\frac{2}{3}:\frac{10}{11}\)

         \(y=\frac{11}{15}\)

\(\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+\frac{3}{9\cdot11}+...+\frac{3}{2013\cdot2015}\)

\(=\frac{3}{2}\cdot\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(=\frac{3}{2}\cdot\left(\frac{1}{5}-\frac{1}{2015}\right)=\frac{3}{2}\cdot\frac{402}{2015}-\frac{603}{2015}\)

Vậy \(\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+\frac{3}{9\cdot11}+...+\frac{3}{2013\cdot2015}=\frac{603}{2015}\)

3/5.7 + 3/7.9 + 3/9.11 + ... 3/2013.2015

= 3/2.( 2/5.7 + 2/7.9 + 2/9.11 + ... + 2/2013.2015)

= 3/2. ( 1/5 - 1/7 + 1/7 - 1/9 +  1/9 - 1/11 + ... + 1/2013 - 1/2015) 

   ~~~  SAU ĐÓ BẠN GẠCH ĐI NHỮNG PHÂN SỐ GIỐNG NHAU NHÁ ~~~

= 3/2. ( 1/5 - 1/2015)

= 3/2. 2010/10075

= 603/4030

Mk chắc chắn cách làm đúng đó!!!

    \(\frac{3}{5\text{×}7}+\frac{3}{7\text{×}9}+.......+\frac{3}{2013\text{×}2015}\)

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

\(=\frac{3}{2}\text{×}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+......+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(=\frac{3}{2}\text{×}\left(\frac{1}{5}-\frac{1}{2015}\right)\)

\(=\frac{3}{2}\text{×}\frac{402}{2015}\)

\(=\frac{603}{2015}\)

\(\left(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+\frac{1}{9x11}\right)xX=\frac{8}{11}\)

\(\Rightarrow\left(\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\right)xX=\frac{16}{11}\)

\(\Rightarrow\left(\frac{5-3}{3x5}+\frac{7-5}{5x7}+\frac{9-7}{7x9}+\frac{11-9}{9x11}\right)xX=\frac{16}{11}\)

\(\Rightarrow\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)xX=\frac{16}{11}\)

\(\Rightarrow\left(\frac{1}{3}-\frac{1}{11}\right)xX=\frac{16}{11}\Rightarrow\frac{8}{33}xX=\frac{16}{11}\)

\(\Rightarrow X=\frac{16}{11}:\frac{8}{33}=\frac{16}{11}x\frac{33}{8}=6\)

Q= 3/3x5 + 3/5x7 + 3/7x9 +...+ 3/47x49

Q= (3/3 -3/5) + (3/5-3/7) + (3/7-3/9)+...+(3/47-3/49)

Q= 3/3 - 3/5 + 3/5  - 3/7 + 3/7 -  3/9 + ... + 3/47 - 3/49

Q=3/3 - 3/49

Q= 46/49

72/49

\(\frac{23}{49}\)