\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

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

\(=\frac{1}{2}.\frac{100}{101}\)

\(=\frac{50}{101}\)

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{99\cdot101}\)

\(=2\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{99\cdot101}\right)\)

\(=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{99\cdot101}\)

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

\(=\frac{1}{1}-\frac{1}{101}=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)

ta có : 2S=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

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

          2S=\(\frac{1}{1}-\frac{1}{101}\)

      2S+\(\frac{1}{101}\)= \(\frac{1}{1}-\frac{1}{101}+\frac{1}{101}\)

      2S+\(\frac{1}{101}\)=1

ok

a, \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{49.50}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\frac{1}{5}-\frac{1}{50}=\frac{9}{50}\)

b, \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)

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

\(=1-\frac{1}{101}=\frac{100}{101}\)

\(\frac{1}{5\times6}+\frac{1}{6\times7}+\frac{1}{7\times8}+...+\frac{1}{49\times50}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\frac{1}{5}-\frac{1}{50}=\frac{9}{50}\)

~ Hok tốt ~

trl/

a) 1/5x6 + 1/6x7 +..+1/49x50=9/50

b?tuowg tự

nhtp

Theo công thức là ra nhé=))

chịu

A=1x3x(5+1) + 3x5x(7-1) +5x7x(9-3) +...+ 99x101x(103-97)

6A=3+ 1x3x5 +3x5x7-1x3x5 + 5x7x9 -3x5x7 +....+99x101x103 - 97x99x101

6A=3+99x101x103=1019703

vậy = 1019703

nếu sai chỗ nào thì sửa hộ mk vs

Ta có:

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

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

\(=1-\frac{1}{101}=\frac{100}{101}< 1\)

Vậy \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}< 1\)

Đặt biểu thức là A 

Ta có A = (3-1)1x3 + (5-3)/3x5+..........+(101-99)/101x99

            =3/1x3 - 1/1*3 + 5/3x5 - 3/3x5 + ...........+ 101/99x101 - 99/101x99

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

              = 1 -1/101 < 1 (Điều phải chứng minh)