\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}\)

\(=\frac{1}{1}-\frac{1}{11}\)

\(=\frac{10}{11}\)

\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

\(=1-\frac{1}{11}\)

\(=\frac{10}{11}\)

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

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

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

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

\(Q=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

     \(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\)\(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

       \(=\frac{1}{3}-\frac{1}{11}=\frac{8}{33}\)

\(Q=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

\(Q=\frac{1}{3}+0+0+0-\frac{1}{11}\)

\(Q=\frac{11}{33}-\frac{3}{33}=\frac{8}{33}\)

\(Q=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}=\frac{1}{3}-\frac{1}{11}=\frac{8}{33}\)

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

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

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

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)

\(=\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+...+\frac{2}{37}-\frac{2}{39}\)

\(=\frac{2}{3}-\frac{2}{39}\)

\(=\frac{8}{13}\)

Ta có:

\(\frac{2}{3.5}=\frac{1}{3}-\frac{1}{5}\)

\(\frac{2}{5.7}=\frac{1}{5}-\frac{1}{7}\)

\(\frac{2}{7.9}=\frac{1}{7}-\frac{1}{9}\)

\(......................................\)

\(\frac{2}{37.39}=\frac{1}{37}-\frac{1}{39}\)

nên \(C=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)

\(C=\frac{1}{3}-\frac{1}{39}=\frac{4}{13}\)

2/3.5 + 2/5.7 + 2/7.9 + ... + 2/41.43

= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/41 - 1/43

= 1/3 - 1/43

= 40/129

ỦNG HỘ NHA

2/3.5 + 2/5.7 + 2/7.9 +......+ 2/41.43

= 1/3-1/5 + 1/5-1/7 + 1/7-1/9 +.....+ 1/41-1/43

= 1/3-1/43

= 40/129.

40/129 chuẩn 100% luôn

Giải :

\(N=\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)

=> \(N=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)

=> \(N=2.\left(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{101}\right)\right)\)

=> \(N=2.\left(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\right)\)

=> \(N=\frac{98}{303}\)

N=1/2x(1/3-1/5+1/5-1/7+.....+1/99-1/101)

N=1/2x(1/3-1/101)

N=1/2x98/303

N=49/303

Ta có công thức \(\frac{a}{b.c}=\frac{a}{c-b}.\left(\frac{1}{b}-\frac{1}{c}\right)\)

Dựa vào công thức ta có :

\(\frac{2}{3.5}=\frac{2}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)\)

\(\frac{2}{5.7}=\frac{2}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)\)

........................................

\(\frac{2}{99.101}=\frac{2}{2}.\left(\frac{1}{99}-\frac{1}{101}\right)\)

\(\Rightarrow\)\(N=1.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{99}-\frac{1}{101}\right)\)

\(\Leftrightarrow\)\(N=\frac{1}{3}-\frac{1}{101}\)

\(\Rightarrow N=\frac{98}{303}\)

Ai thấy đúng thì ủng hộ nha !!!!

1-1/3-1/65

\(A=1-\frac{2}{3.5}-\frac{2}{5.7}-\frac{2}{7.9}-...-\frac{2}{63.65}\)

\(A=1-\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{63-65}\right)\)

\(A=1-\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{63}-\frac{1}{65}\right)\)

\(A=1-\left(\frac{1}{3}-\frac{1}{65}\right)\)

\(A=1-\frac{62}{195}\)

\(A=\frac{133}{195}\)

Đề bài sai 

\(A=\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.100}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{99}-\frac{1}{100}\)

\(A=\frac{1}{3}-\frac{1}{100}=\frac{97}{300}\)

A=1/3-1/5+1/5-1/7+...+1/99-1/101 là 2/99.101 nhé bạn mình làm nhiều rồi có lẽ bạn ghi đề sai

A=98/303