\(\frac{4}{4.6}+\frac{4}{6.8}+\frac{4}{8.10}+...+\frac{4}{28.30}\)

\(=2.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+...+\frac{2}{28.10}\right)\)

\(=2.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+...+\frac{1}{28}-\frac{1}{30}\right)\)

\(=2.\left(\frac{1}{4}-\frac{1}{30}\right)=2.\left(\frac{15}{60}-\frac{2}{60}\right)=2.\frac{13}{60}=\frac{26}{60}=\frac{13}{30}\)

hello 

a)   \(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+\frac{1}{8\cdot10}\)

\(=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)

\(=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(=\frac{1}{2}\cdot\frac{2}{5}=\frac{2}{10}=\frac{1}{5}\)

b)   \(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}\)

\(=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}\)

\(=1-\frac{1}{17}=\frac{16}{17}\)

hok tốt ...

a)\(A=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+\frac{1}{8\cdot10}\)

\(2A=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+\frac{2}{8\cdot10}\)

\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)

\(A=\frac{2}{5}\cdot\frac{1}{2}=\frac{1}{5}\)

b)\(B=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}=1-\frac{1}{17}=\frac{16}{17}\)

a) 1/2.4 + 1/4.6 + 1/6.8 + 1/8.10

= 2/2.2.4 + 2/2.4.6 + 2/2.6.8 + 2/2.8.10 ( nhân cả tử và mẫu với 2)

= 1/2 .( 2/2.4 + 2/4.6 + 2/6.8 + 2/8.10 )

= 1/2 .(1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/10)

= 1/2.(1/2 - 1/10)

= 1/2.( 5/10 - 1/10) = 1/2.4/10 = 2/10 = 1/5

b) 4/1.5+ 4/5.9 + 4/ 9.13 + 4/13.17

= 1- 1/5 + 1/5 - 1/9 + 1/9 - 1/13 + 1/13 - 1/17

= 1- 1/17

= 16/17 

trong sách nâng cao và phát triển 6 đó bạn

=5/2(1/4-1/6+1/6-1/8+...+1/208-1/300)

=5/2(1/4-1/300)

=5/2.37/150=37/60

Q=1/4(1.4/2.3+2.5/3.4+3.6/4.5+...+48.51/49.50)

=1/4(2.3−2/2.3+3.4−2/3.4+4.5−2/4.5+...+49.50−2/49.50)

=1/4(1− 2/2.3+ 1− 2/3.4+ 1− 2/4.5+...+1− 2/49.50)

=1/4[48−2(1/2.3+1/3.4+...+1/49.50)]

=1/4[48−2(1/2−1/3+1/3−1/4+...+1/49−150)]

=14[48−2(1/2−1/50)]=294/25

\(S=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)

\(S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\)

\(S=\frac{1}{2}-\frac{1}{10}\)

\(S=\frac{2}{5}\)

S = 1 / 2.4 + 1/ 4.6 + 1 / 6.8 + 1 / 8.10

2S = 2 / 2.4 + 2 / 4.6 + 2/ 6.8 + 2 / 8.10

2S = 1 2 - 1 / 4 + 1 / 4 - 1 / 6 + 1 / 6 - 1 / 8 + 1 / 8 - 1 / 10

2S = 1 / 2 - 1 / 10

2S = 2 / 5 

  S = 2 / 5 : 2

  S = 1 / 5 

\(\frac{5}{4\cdot6}+\frac{5}{6\cdot8}+\frac{5}{8\cdot10}+...+\frac{5}{298\cdot300}\) 

\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{298}-\frac{1}{300}\right)\) 

\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{300}\right)\) 

\(=\frac{5}{2}\cdot\frac{37}{150}\) 

\(=\frac{37}{60}\)

\(\frac{5}{4.6}+\frac{5}{6.8}+\frac{5}{8.10}+...+\frac{5}{298.300}\)

= \(\frac{5}{2}.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+...+\frac{2}{298.300}\right)\)

= \(\frac{5}{2}.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{298}-\frac{1}{300}\right)\)

= \(\frac{5}{2}.\left(\frac{1}{4}-\frac{1}{300}\right)\)

= \(\frac{5}{2}.\frac{37}{150}\)

= \(\frac{37}{60}\)

Ta đặt biểu thức trên là A

\(\frac{1}{2}\)A=\(\frac{2}{4.6}\)+\(\frac{2}{6.8}\)+...........+\(\frac{2}{298.300}\)

\(\frac{1}{2}\)A=1/4-1/6+1/6-1/8+..............+1/298-1/300

\(\frac{1}{2}\)A=1/4-1/300

1/2A=74/300

A=74/300:1/2

A=37/75

\(S=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)

\(2S=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\)

\(2S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\)

\(2S=\frac{1}{2}-\frac{1}{10}\)

\(2S=\frac{2}{5}\)

\(S=\frac{2}{5}:2\)

\(S=\frac{1}{5}\)

S = \(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)

=> 2S = \(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\)

=> 2S = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\)

=> 2S = \(\frac{1}{2}-\frac{1}{10}=\frac{5}{10}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\)

=> S = \(\frac{2}{5}:2=\frac{2}{5}x\frac{1}{2}=\frac{1}{5}\)

\(S=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)

\(\Rightarrow2S=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\)

\(\Rightarrow2S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\)

\(\Rightarrow2S=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)

\(\Rightarrow S=\frac{2}{5}:2=\frac{2}{5}.\frac{1}{2}=\frac{1}{5}\)