12.T=2.6.12+6.10.12+10.14.12+...+102.106.12=

=2.6.(10+2)+6.10.(14-2)+10.14.(18-6)+...+102.106.(110-98)=

=2.2.6+2.6.10-2.6.10+6.10.14-6.10.14+10.14.18-...-98.102.106+102.106.110=

=2.2.6+102.106.110

\(\Rightarrow T=\dfrac{2.2.6+102.106.110}{12}=99112\)

bạn j ơi cho mình hỏi là 102 lấy ở đâu ja <3

Đặt A = 1/2.6 + 1/6.10 + 1/10.14 + ..... + 1/102.106

=> 4A = 4/2.6 + 4/6.10 + 4/10.14 + ..... + 4/102.106

=> 4A = 1/2 - 1/6 + 1/6 - 1/10 + 1/10 - 1/14 + ... + 1/102 - 1/106

=> 4A = 1/2 - 1/106

=> 4A = 26/53

=> A = 13/106

~Study well~

#QASJ

\(\frac{1}{2.6}+\frac{1}{6.10}+...+\frac{1}{102.106}\)

\(=\frac{1}{4}.\left(\frac{4}{2.6}+\frac{4}{6.10}+...+\frac{4}{102.106}\right)\)

\(=\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+...+\frac{1}{102}-\frac{1}{106}\right)\)

\(=\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{106}\right)\)

\(=\frac{1}{4}.\frac{26}{53}\)

\(=\frac{13}{106}\)

Đặt \(B=\frac{1}{2.6}+\frac{1}{6.10}+\frac{1}{10.14}+...+\frac{1}{102.106}\)

\(\Leftrightarrow4B=\frac{4}{2.6}+\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{102.106}\)

\(\Leftrightarrow4B=\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{102}-\frac{1}{106}\)

\(\Leftrightarrow4B=\frac{1}{2}-\frac{1}{106}=\frac{26}{53}\)

\(\Leftrightarrow B=\frac{26}{53}:4=\frac{13}{106}\)

Vậy ...................

~ Hok tốt ~

Sửa đề: \(D=\dfrac{3}{2\cdot6}+\dfrac{3}{6\cdot10}+\dfrac{3}{10\cdot14}+...+\dfrac{3}{26\cdot30}\)

Ta có: \(D=\dfrac{3}{2\cdot6}+\dfrac{3}{6\cdot10}+\dfrac{3}{10\cdot14}+...+\dfrac{3}{26\cdot30}\)

\(=\dfrac{3}{4}\left(\dfrac{4}{2\cdot6}+\dfrac{4}{6\cdot10}+\dfrac{4}{10\cdot14}+...+\dfrac{4}{26\cdot30}\right)\)

\(=\dfrac{3}{4}\left(\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{1}{26}-\dfrac{1}{30}\right)\)

\(=\dfrac{3}{4}\left(\dfrac{1}{2}-\dfrac{1}{30}\right)\)

\(=\dfrac{3}{4}\cdot\dfrac{28}{60}\)

\(=\dfrac{21}{60}=\dfrac{7}{20}\)

a) 2/2.5 + 2/5.8 + 2/8.11 + ... + 2/x(x+3) = 7/23

3/2.5 + 3/5.8 + 3/8.11 + ... + 3/x(x+3) = 21/46

1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/x - 1/x+1 = 21/46

1/2 - 1/x+1 = 21/46

=> 1/x+1 = 1/23

=> x + 1 = 23

=> x = 22

Vậy x = 22.

b) 3/4 . x - 1/5 = 7/4 . x + 11/5

3/4 . x - 7/4 . x = 1/5 + 11/5

x (3/4 - 7/4) = 12/5

-x = 12/5

x = -12/5

Vậy x = -12/5.

   \(\dfrac{3}{2.6}\) + \(\dfrac{3}{6.10}\) + \(\dfrac{3}{10.14}\)

=  \(\dfrac{3}{4}\).(\(\dfrac{4}{2.6}\) + \(\dfrac{4}{6.10}\) + \(\dfrac{4}{10.14}\))

= \(\dfrac{3}{4}\).(\(\dfrac{1}{2}-\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{14}\))

= \(\dfrac{3}{4}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{14}\))

= \(\dfrac{3}{4}\). \(\dfrac{3}{7}\)

= \(\dfrac{9}{28}\)

B = \(\dfrac{4}{1.3.5}\) + \(\dfrac{4}{3.5.7}\) + \(\dfrac{4}{5.7.9}\)

B = \(\dfrac{1}{1.3}\) - \(\dfrac{1}{3.5}\) + \(\dfrac{1}{3.5}\) - \(\dfrac{1}{5.7}\) + \(\dfrac{1}{5.7}\) - \(\dfrac{1}{7.9}\)

B = \(\dfrac{1}{1.3}\) - \(\dfrac{1}{7.9}\)

B = \(\dfrac{1}{3}\) - \(\dfrac{1}{63}\)

B =  \(\dfrac{20}{63}\)

C = \(\dfrac{5}{2.4.6}\) + \(\dfrac{5}{4.6.8}\) + \(\dfrac{5}{6.8.10}\)

C = \(\dfrac{5}{4}\).(\(\dfrac{4}{2.4.6}\) + \(\dfrac{4}{4.6.8}\) + \(\dfrac{4}{6.8.10}\))

C = \(\dfrac{5}{4}\).(\(\dfrac{1}{2.4}\) - \(\dfrac{1}{4.6}\) + \(\dfrac{1}{4.6}\) - \(\dfrac{1}{6.8}\) + \(\dfrac{1}{6.8}\) - \(\dfrac{1}{8.10}\))

C = \(\dfrac{5}{4}\).(\(\dfrac{1}{2.4}\) - \(\dfrac{1}{8.10}\))

C = \(\dfrac{5}{4}\).( \(\dfrac{1}{8}\) - \(\dfrac{1}{80}\))

C = \(\dfrac{5}{4}\). \(\dfrac{9}{80}\)

C = \(\dfrac{9}{64}\)