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

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

= \(1-\dfrac{1}{101}\)

=\(\dfrac{100}{101}\) 

\(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{5.7}+...+\dfrac{5}{99.101}\)

=\(\dfrac{5}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99+101}\right)\)

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

=\(\dfrac{5}{2}.\left(1-\dfrac{1}{101}\right)\)

= \(\dfrac{5}{2}-\dfrac{100}{101}\)

= \(\dfrac{305}{202}\)

917749738461936926399639748776398646491639394748947630373937366

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

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

sửa đề nhé:

\(\frac{5}{1\times3}+\frac{5}{3\times5}+...+\frac{5}{99\times101}\)

\(5\times\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+...+\frac{2}{99\times101}\right)\)

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

\(=\frac{5}{2}\times\left(1-\frac{1}{101}\right)\)

\(=\frac{5}{2}\times\frac{100}{101}\)

\(=\frac{250}{101}\)

mong cac ban giup

A=1x3 +3x5 +5x7 +....+99x101

6A=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