B : 7/2 =2/1.3+2/3.5+...+2/99.101

B:7/2=1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101

B:7/2=1-1/101=100/101

B=100/101*7/2=700/202=350/101

B=7/2(2/1.3+2/3.5+ ...+2/99.101)

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

B=7/2(1-1/101)=7/2.100/101=350/101

k nha bạn

B=1/1-1/3+1/3-1/5+1/5-1/7+....+1/99-1/101

B=1-1/101=100/101

vậy B=100/101

7/1.3 + 7/3.5 + 7/5.7 + ... + 7/99.101

= 7.(1/1.3 + 1/3.5 + 1/5.7 + ... + 1/99.101)

= 7/2 . 2 . (1/1.3 + 1/3.5 + 1/5.7 + ... + 1/99.101)

= 7/2 . (2/1.3 + 2/3.5 + 2/5.7 + ... + 2/99.101)

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

= 7/2 . (1 - 1/101)

= 7/2 . 100/101

= 350/101

\(\frac{7}{1.3}+\frac{7}{3.5}+...+\frac{7}{99.101}\)

\(=7\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right)\)

\(=\)\(\frac{7}{2}.2.\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right)\)

\(=\)\(\frac{7}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)

=\(\frac{7}{2}x\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{99}-\frac{1}{100}\right)\)

=\(\frac{7}{2}x\left(\frac{1}{3}-\frac{1}{100}\right)\)

=\(\frac{7}{2}\)x\(\frac{97}{300}\)

=\(\frac{679}{600}\)

a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

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

\(=1-\frac{1}{101}\)

\(=\frac{100}{101}\)

b) \(\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\)

\(=7.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\right)\)

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

\(=\frac{7}{7}\left(1-\frac{1}{101}\right)\)

\(=\frac{100}{101}\)

mình giống như bạn Phạm Hữu Đang

tham the 

có giỏi thì làm một câu xem nào

đơn giản à

\(\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+....+\frac{7}{99.101}\)

\(=\frac{7}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\right)\)

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

\(=\frac{7}{2}\left(1-\frac{1}{101}\right)=\frac{7}{2}.\frac{100}{101}=\frac{350}{101}\)

= \(\frac{350}{101}\)

=1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101

=1-1/101

=100/101

k cho mình nha

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)

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

\(=\frac{1}{2}.\left(1-\frac{1}{101}\right)=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)

TA CÓ \(\frac{1}{1.3}+\frac{1}{3.5}+.....+\frac{1}{99.101}\)

              \(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{101}\)

                \(=\frac{1}{1}-\frac{1}{101}\)

                  \(=\frac{100}{101}\)

\(A=\frac{7}{1.3}+\frac{7}{3.5}+.............+\frac{7}{99.101}\)

\(=\frac{7}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+........+\frac{2}{99.101}\right)\)

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

\(=\frac{7}{2}.\left(1-\frac{1}{101}\right)\)

\(=\frac{7}{2}.\frac{100}{101}\)

\(=\frac{350}{101}\)

Tính :

a) \(M=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

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

\(=1-\frac{1}{101}\)

\(=\frac{100}{101}\)

b) \(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

\(=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)

\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)

\(=7.\frac{3}{35}\)

\(=\frac{3}{5}\)

c) \(B=\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)

\(=\frac{1}{2}.\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)\)

\(=\frac{1}{2}.\frac{2}{75}\)

\(=\frac{1}{75}\)

thanks

M=2/1.3+2/3.5+2/5.7+...+2/99.101

    =1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101

    =1-1/101

    =101/101-1/101

 M = 99/101

A=7/10.11+7/11.12+7/12+13+7/69/70 ( sai đề )

 B= 1/25.27+1/27.29+1/29.31+...+1/73.75

   =1/2.(2/25.27+2/27.29+2/29.31+...+2/73.75

   =1/2.(1/25-1/27+1/27-1/29+1/29-1/31+...+1/73-1/75

   =1/2.(1/25-1/75)

   =1/2.75/100

   =3/8