99/200

\(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+....+\frac{1}{98\cdot100}\)

\(=\frac{1}{2}\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+.......+\frac{2}{98\cdot100}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+......+\frac{1}{98}-\frac{1}{100}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{49}{200}\)

Đặt A = 1/2.4 + 1/4.6 + .... +1/98.100

\(2A=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\)

       \(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\)

        \(=\frac{1}{2}-\frac{1}{100}\)

         \(=\frac{49}{100}\)

\(\Rightarrow A=\frac{49}{100}\div2=\frac{49}{200}\)

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

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

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

2)\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)

\(=2\times\frac{502}{1005}\)

\(=\frac{1004}{1005}\)

tự làm tiếp nhé

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

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

2.= \(2\cdot\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\right)\)

   = \(2\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

   = \(2\cdot\left(\frac{1}{2}-\frac{1}{2010}\right)\) = \(2\cdot\frac{502}{1005}\) = \(\frac{1004}{1005}\)

a)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)

b)\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+..+\frac{4}{2008.2010}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1004.1005}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1004}-\frac{1}{1005}=1-\frac{1}{1005}=\frac{1004}{1005}\)

c)\(\frac{1}{2}.\left(\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+...+\frac{1}{9800}\right)=\frac{1}{4}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{98.100}\right)=\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{49}{400}\)

A=1/2.4+1/4.6+........+1/100.102

A=1/2-1/4+1/4-1/6+.......+1/100-1/102

A=1/2-1/102

A=51/102-1/102

A=50/102

A=25/51

\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{18}-\frac{1}{20}\)

\(A=\frac{1}{2}-\frac{1}{20}\)

\(A=\frac{10}{20}-\frac{1}{20}\)

\(A=\frac{9}{20}\)

\(\frac{1}{2.4}+\frac{1}{4.6}+....+\frac{1}{2016.2018}\)

\(=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2016.2018}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2018}\right)=\frac{1}{2}.\frac{504}{1009}=\frac{252}{1009}\)

mơn bn nha

\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2016.2018}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2016}-\frac{1}{2018}\)

\(=\frac{1}{2}+\left(-\frac{1}{4}+\frac{1}{4}\right)+\left(-\frac{1}{6}+\frac{1}{6}\right)+...+\left(-\frac{1}{2016}+\frac{1}{2016}\right)-\frac{1}{2018}\)

\(=\frac{1}{2}+0+0+0+...+0-\frac{1}{2018}\)

\(=\frac{1}{2}-\frac{1}{2018}\)

\(=\frac{1009}{2018}-\frac{1}{2018}\)

\(=\frac{1008}{2018}=\frac{504}{1009}\)

Ai thấy tớ đúng k nha

Gọi tổng cần tính là \(A\)

Ta có: \(A=\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{38.40}\)

\(\Rightarrow2A=\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{38.40}\)

\(\Rightarrow2A=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{38}-\dfrac{1}{40}\)

\(\Rightarrow2A=\dfrac{1}{2}-\dfrac{1}{40}=\dfrac{19}{40}\)

\(\Rightarrow A=\dfrac{\dfrac{19}{40}}{2}=\dfrac{19}{80}\)

1/ 2.4 + 1/4.6 + ...+1/18.20

= 1/2 - 1/4 + 1/4 -1/6 + .... + 1/18.20

trừ hết đi cho nhau cuối cùng:

= 1/2 - 1/20 = 9/20