Nhân cả tổng với 2/2.

\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+....+\frac{5}{48.50}\)

\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)

\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)

\(=\frac{5}{2}.\frac{12}{25}=\frac{6}{5}\)

\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+...+\frac{5}{48.50}\)

\(=\frac{2}{5}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{48.50}\right)\)

\(=\frac{2}{5}.\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{50-48}{48.50}\right)\)

\(=\frac{2}{5}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)

\(=\frac{2}{5}.\left(\frac{1}{2}-\frac{1}{50}\right)\)

\(=\frac{2}{5}.\frac{12}{25}\)

\(=\frac{24}{125}\)

\(A=\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{48.50}\)

\(A=\frac{5}{2}\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{48.50}\right)\)

\(A=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{50}\right)\)

\(A=\frac{6}{5}\)

=\(\frac{1}{5}.\left(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{48}-\frac{1}{50}\right)\right)\)

=\(\frac{1}{5}.\left(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\right)\)

=\(\frac{1}{5}.\left(\frac{1}{2}.\frac{12}{25}\right)\)

=\(\frac{1}{5}.\frac{6}{25}=\frac{6}{125}\)

Vậy \(A=\frac{6}{125}\)

A = (5/2 - 5/4) + (5/4-5/6) +....+(5/48-5/50)

A = 5/2 + ( 5/4-5/4) +(5/6-5/6)+....+(5/48-5/48)+5/50

A = 5/2-5/50

A = 125/50-5/50

A = 120/50 = 12/5

\(\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{98.100}\)

= \(\frac{5}{2}-\frac{5}{4}+\frac{5}{4}-\frac{5}{6}+...+\frac{5}{98}-\frac{5}{100}\)

= \(\frac{5}{2}-\frac{5}{100}\)

= \(\frac{49}{50}\)

\(Q=\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{98.100}\)

    \(=5\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)

    \(=\frac{5}{2}.2.\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)

    \(=\frac{5}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\right)\)

    \(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)

    \(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{5}{2}.\frac{49}{100}=\frac{49}{40}\)

\(\Rightarrow Q=\frac{49}{40}\)

\(B=\frac{3}{2.4}-\frac{5}{4.6}+\frac{7}{6.8}-\frac{9}{8.10}+...+\frac{2019}{2018.2020}\)

\(B=\frac{3}{2.1.2.2}-\frac{5}{2.2.2.3}+\frac{7}{2.3.2.4}-\frac{9}{2.4.2.5}+...+\frac{2019}{2.1009.2.1010}\)

\(B=\frac{1}{4.}.\left(\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+...+\frac{2019}{1009.1010}\right)\)

\(B=\frac{1}{4.}.\left(\frac{3}{1}-\frac{3}{2}-\frac{5}{2}+\frac{5}{3}+\frac{7}{3}-\frac{7}{4}-\frac{9}{4}+\frac{9}{5}+...+\frac{2019}{1009}-\frac{2019}{1010}\right)\)

\(B=\frac{1}{4.}.\left(\frac{3}{1}-4+4-4+4-...+4-\frac{2019}{1010}\right)\)

\(B=\frac{1}{4.}.\left(\frac{3}{1}-\frac{2019}{1010}\right)=\frac{1011}{4040}\)

\(=\left(1+3+5+...+99+101\right)-\left(2+4+6+...98+100\right)\)
Thấy từ 1 đến 100 có (101-1)/2+1=51
=> 1+3+5+....+99+100=(1+101)x50/2=2601
Từ 2 đến 100 có (102-2)/2+1=50
=> 2+4+...+98+100=(2+100)X50/2=2550
=> D=2601-2550=51

2/2*4 + 2/4*6 + 3/6*8 + ... + 2/38*50

= 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + .... + 1/38 - 1/50

= 1/2 - 1/50

= 24/50

=  12/25

Mk có cách khác câu d nè

\(D=\left(1-2\right)+\left(3-4\right)+...+\left(99-100\right)+101\)

     \(=-1-1-1-...-1+101\)(có 50 số -1)

     \(=-1\times50+101\)

      \(=51\)

chúc bn hok tốt