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

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

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

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

\(B=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)

\(B=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)

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

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

\(B=\frac{5}{2}.\frac{100}{101}\)

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

\(\frac{2}{1.2}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.101}\)

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

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

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

Bạn giúp mk nốt b được ko?

Vậy mà cũng gọi là trả lời 

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

\(\frac{2}{1\cdot3}=\frac{3-1}{1\cdot3}=\frac{3}{1\cdot3}-\frac{1}{1\cdot3}=\frac{1}{1}-\frac{1}{3}=1-\frac{1}{3}\)

\(\frac{2}{3\cdot5}=\frac{5-3}{3\cdot5}=\frac{5}{3\cdot5}-\frac{3}{3\cdot5}=\frac{1}{3}-\frac{1}{5}\)

....

\(\frac{2}{99\cdot101}=\frac{101-99}{99\cdot101}=\frac{101}{99\cdot101}-\frac{99}{99\cdot101}=\frac{1}{99}-\frac{1}{101}\)

\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)

\(\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+\frac{5}{5\cdot7}+...+\frac{5}{99\cdot101}\)

=\(\frac{5}{2}\cdot\frac{2}{1\cdot3}+\frac{5}{2}\cdot\frac{2}{3\cdot5}+\frac{5}{2}\cdot\frac{2}{5\cdot7}+...+\frac{5}{2}\cdot\frac{2}{99\cdot101}\)

=\(\frac{5}{2}\cdot\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right]\)

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

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

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

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

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

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

gạch gạch gạch gạch ... gạch gạch

= 1 - 1 / 101

= 100 / 101

Link nè lên google search nha!

https://olm.vn/hoi-dap/question/162533.html

A = \(\frac{1}{1\cdot3}\)+  \(\frac{1}{3.5}\)+ \(\frac{1}{5.7}\)+  ..... + \(\frac{1}{99.101}\)

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

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

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

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

   Thấy đúng thì cho mình một k nha!!!

A=1/2*100/101=50/101

cái này bạn mở sách bồi dưỡng toán ra trang gần cuối là thấy ngay ấy mà

\(A=\dfrac{4}{1\cdot3}+\dfrac{4}{3\cdot5}+\dfrac{4}{5\cdot7}+...+\dfrac{4}{99\cdot101}\)

\(A=2\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{4}{99\cdot101}\right)\)

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

\(A=2\cdot\left(1-\dfrac{1}{101}\right)\)

\(A=2\cdot\dfrac{100}{101}\)

\(A=\dfrac{200}{101}\)