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

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

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

\(=\frac{5}{2}\cdot\frac{37}{150}\) 

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

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

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

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

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

= \(\frac{5}{2}.\frac{37}{150}\)

= \(\frac{37}{60}\)

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

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

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

\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{300}\right)=\frac{5}{2}.\frac{37}{150}=\frac{37}{60}\)

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

\(=\frac{5}{2}\cdot\left(\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{298\cdot300}\right)\)

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

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

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

\(\frac{5}{4.6}+\frac{5}{6.8}+\frac{5}{8.10}+...+\frac{5}{198.200}\)

\(=\frac{5}{2}\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+...+\frac{2}{198.200}\right)\)

\(=\frac{5}{2}\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)

\(=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)

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

\(=\frac{5}{2}\left(\frac{50}{100}-\frac{1}{100}\right)\)

\(=\frac{5}{2}.\frac{49}{100}\)

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

Ta sẽ tách 5 ra ngoài

Bạn Vũ Đình phước ơi sao bạn ko tách số 5 ở tử rồi nhân thêm 5 ở phân số.

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

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

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

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

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

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

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

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

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

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

trong sách nâng cao và phát triển 6 đó bạn

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

b) Ta có: \(S=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{298\cdot300}\)

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

\(=\frac{1}{2}-\frac{1}{300}=\frac{149}{300}< \frac{200}{300}=\frac{2}{3}\)

hay \(S< \frac{2}{3}\)(1)

Ta có: \(\frac{1}{101}>\frac{1}{102}>\frac{1}{103}>...>\frac{1}{300}\)

nên \(\left(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\right)+\left(\frac{1}{201}+\frac{1}{202}+\frac{1}{203}+...+\frac{1}{300}\right)>\left(\frac{1}{200}+\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\right)+\left(\frac{1}{300}+\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}\right)\)(vì mỗi ngoặc trên đều có 100 phân số có tử là 1)

\(\Leftrightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{300}>\frac{1}{200}\cdot100+\frac{1}{300}\cdot100\)

\(\Leftrightarrow Q>\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)

mà \(\frac{5}{6}>\frac{4}{6}=\frac{2}{3}\)

nên \(Q>\frac{2}{3}\)

hay \(\frac{2}{3}< Q\)(2)

Từ (1) và (2) suy ra S<Q