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

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

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

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

tk cho minh nhe >.<

Ta có: \(\frac{2}{3}\times\left(\frac{3}{5.7}+\frac{3}{7.9}+.....+\frac{3}{59.61}\right):\frac{2}{3}\)

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

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

\(\Rightarrow\left(\frac{1}{5}-\frac{1}{61}\right):\frac{2}{3}=\frac{56}{305}:\frac{2}{3}=\frac{84}{305}\)

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

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

= \(\frac{3}{2}\). ( \(\frac{1}{5}\)- \(\frac{1}{61}\))  = \(\frac{3}{2}\). \(\frac{61-5}{305}\)= \(\frac{3}{2}\). \(\frac{56}{305}\)

= \(\frac{3.28}{1.305}\)= \(\frac{82}{305}\)

\(=\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+...+\frac{4}{59.61}\)

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

\(=2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{59}-\frac{1}{61}\right)\)

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

\(=2.\left(\frac{36}{505}\right)\)

\(=\frac{72}{505}\)

TK nha !!

Ta có : \(\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+....+\frac{4}{59.61}\)

\(=2\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+.....+\frac{2}{59.61}\right)\)

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

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

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

Mk nhân lộn ở cái phần \(2.\left(\frac{1}{5}-\frac{1}{61}\right)\)

\(=2.\left(\frac{56}{305}\right)\)

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

Tk nha cảm ơn !!

ko co de]

\(\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+...+\frac{3}{59.61}\) 

\(=\)\(\frac{3}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{59.61}\right)\)

\(=\)\(\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{59}-\frac{1}{61}\right)\)

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

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

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

Chúc bạn học tốt ~ 

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

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

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

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

\(=\frac{3}{2}.\left(\frac{61}{305}-\frac{5}{305}\right)\)

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

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

\(S=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{20}}\)

\(\Rightarrow2S=1+\frac{1}{2}+...+\frac{1}{2^{19}}\)

\(\Rightarrow2S-S=1-\frac{1}{2^{20}}\)

\(\Rightarrow S=1-\frac{1}{2^{20}}\)

Chúc bạn học tốt !!! 

A = 2/3*5 + 2/5*7 + 2/7*9 + ... + 2/97*99

A = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/97 - 1/99

A = 1/3 - 1/99

A = 32/99

\(A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

\(A=\frac{1}{3}-\frac{1}{99}\)

\(A=\frac{32}{99}\)

\(A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

\(A=\frac{1}{3}-\frac{1}{99}\)

\(A=\frac{32}{99}\)

a.\(\frac{3\cdot4\cdot7}{12\cdot8\cdot9}\)= \(\frac{3\cdot4\cdot7}{3\cdot4\cdot8\cdot9}\)= \(\frac{7}{72}\) 

b. \(\frac{4\cdot5\cdot6}{12\cdot10\cdot8}\)= \(\frac{4\cdot5\cdot2\cdot3}{3\cdot4\cdot5\cdot2\cdot8}\)= \(\frac{1}{8}\) 

c.\(\frac{5\cdot6\cdot7}{12\cdot14\cdot15}\)= \(\frac{5\cdot6\cdot7}{2\cdot6\cdot2\cdot7\cdot3\cdot5}\)= \(\frac{1}{12}\)

a, \(\frac{3.4.7}{12.8.9}\)= \(\frac{3.4.7}{3.4.8.9}\)= \(\frac{7}{72}\)

b, \(\frac{4.5.6}{12.10.8}\)= \(\frac{4.5.6}{3.4.2.5.8}\)= \(\frac{1}{8}\)

c, \(\frac{5.6.7}{12.14.15}\)= \(\frac{5.6.7}{2.6.2.7.3.5}\)= \(\frac{1}{12}\)

toán 1 sao?

\(\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+...+\frac{5}{53\cdot55}\)

\(=\frac{3}{2}\cdot\left(\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{53\cdot55}\right)\)

\(=\frac{3}{2}\cdot\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{53}-\frac{1}{55}\right)\)

\(=\frac{3}{2}\cdot\left(\frac{1}{5}-\frac{1}{55}\right)\)

\(=\frac{3}{2}\cdot\left(\frac{11}{55}-\frac{1}{55}\right)\)

\(=\frac{3}{2}\cdot\frac{10}{55}\)

\(=\frac{3}{2}\cdot\frac{2}{11}\)

\(=\frac{3}{11}\)

đây mà là lớp 1 ak 

\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)

\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)

\(=\frac{1}{1.3}-\frac{1}{11.13}\)

\(=\frac{1}{3}-\frac{1}{143}\)

\(=\frac{140}{429}\)

\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)

=>\(S=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)

=>\(S=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)

=>\(S=\frac{1}{2}.\left(1-\frac{1}{9}\right)-\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{10}\right)\)

=>\(S=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)

=>\(S=\frac{4}{9}-\frac{1}{5}\)

=>\(S=\frac{11}{45}\)

lê chí cường dung