\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)

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

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

Tự tính

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

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

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

32/99

k với nghe bạn

và chúc chueeuf nay thi tốt

Ta có: \(M=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+......+\frac{2}{97.99}\)

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

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

        \(M=\frac{32}{99}\)

\(M=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97\cdot99}\)

\(M=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

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

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

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

đúng nha Miyuki

M=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

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

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

TICK ỦNG HỘ NHA

\(M=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{97}-\frac{1}{99}\right).\)

 \(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)=\frac{1}{2}x\frac{32}{99}=\frac{32}{198}\)

bn tự rút gọn nha mk mới làm tắt đó

Ta có : \(M=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

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

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

\(=\frac{33}{99}-\frac{1}{99}\)

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

32/99

\(\frac{2}{3.5}+\frac{2}{5.7}+.................+\frac{2}{97.99}\)

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

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

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

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

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

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

Ta có: \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

 \(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

 \(=\frac{1}{3}-\frac{1}{99}\)

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

F = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

F = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

F = \(\frac{1}{3}-\left(\frac{1}{5}-\frac{1}{5}\right)-\left(\frac{1}{7}-\frac{1}{7}\right)-\left(\frac{1}{9}-\frac{1}{9}\right)-...-\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)

F = \(\frac{1}{3}-\frac{1}{99}\)

F = \(\frac{32}{99}\)

\(F=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97\cdot99}\)

\(\Rightarrow F=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{97}-\frac{1}{99}\)

\(\Rightarrow F=\frac{1}{3}-\frac{1}{99}\)

\(\Rightarrow F=\frac{32}{99}\)

\(F=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(F=1.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(F=1.\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(F=1.\frac{32}{99}\)

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

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

= 1/3 -1/99

=32/99

tích cho mình nha

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

=1/3-1/99

=32/99

a) \(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)

\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)

\(=\frac{1}{5}-\frac{1}{12}\)

\(=\frac{12}{60}+\frac{-5}{60}\)

\(=\frac{7}{60}\)

b) \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=\frac{2}{3}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\right)\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

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

\(=\frac{33}{99}-\frac{1}{99}\)

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

a) \(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{132}=\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-...-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)

\(=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)

b) \(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

b) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=2.\left(1-\frac{1}{99}\right)\)

\(=2.\frac{98}{99}\)

\(=\frac{196}{99}=1\frac{97}{99}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(B=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\)

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

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

\(=\frac{98}{99}\)

A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)

=>\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

=>\(\frac{1}{1}-\frac{1}{100}\)

=>\(\frac{99}{100}\)

B=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{97.99}\)

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

=>\(\frac{1}{1}-\frac{1}{99}\)

=>\(\frac{98}{99}\)