a) 6B = 2.4.6 + 4.6.(8-2) + 6.8.(10-4) + ... + 18.20.(22-16)

    6B  = 2.4.6 + 4.6.8 - 2.4.6 + 6.8.10 - 4.6.8 +...+ 18.20.22 - 16.18.20

     6B = 18.20.

      B = (18.20.22) : 6

      B = 1320
Mấy bài kia tương tự, cần giải luôn không bạn? Nhưng hơi mất thời gian

\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

\(S=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}=\frac{1}{8}-\frac{1}{10}\right)\)

\(S=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}\right)\)

\(S=\frac{1}{2}.\left(\frac{58}{45}\right)\)

\(S=\frac{29}{45}\)

S =1/1.3-1/2.4+1/3.5-1/4.6+1/5.7 - 1/6.8+1/7.9-1/8.10

\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{7.9}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{8.10}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{7.9}\right)+\frac{1}{2}\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{8.10}\right)\)

\(=\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\right)+\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\right)\)

\(=\left(1-\frac{1}{9}\right)+\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(=\frac{8}{9}+\frac{2}{5}\)

\(=\frac{58}{45}\)

viết đề hẳn hoi đi đề thì xấu còn bày đặt làm càn

\(A=\) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}\)

\(A=\frac{49}{50}\)

\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.50}\)

A= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)

A = \(\frac{1}{1}-\frac{1}{51}=\frac{50}{51}\)

\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\frac{1}{2.4}-\frac{1}{4.6}-\frac{1}{6.8}-\frac{1}{8.10}\)

\(S=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)

\(S=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)

\(S=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)

\(S=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(S=\frac{1}{2}\left(1-\frac{1}{9}-\frac{1}{2}+\frac{1}{10}\right)\)

\(S=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)

\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

\(S=\frac{1}{2}.\left[\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\right]\)

\(S=\frac{1}{2}.\left[\left(1-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)-\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\right)\right]\)

\(S=\frac{1}{2}.\left[\left(1-\frac{1}{9}\right)-\left(\frac{1}{2}-\frac{1}{10}\right)\right]\)

\(S=\frac{1}{2}.\left(\frac{8}{9}-\frac{2}{5}\right)\)

\(S=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)

\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

=> \(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)

=> \(S=\frac{1}{2}\left(1-\frac{1}{3}+.....+\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+....+\frac{1}{8}-\frac{1}{10}\right)\)

=> \(S=\frac{1}{2}\left(1-\frac{1}{9}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)

=> \(S=\frac{1}{2}.\frac{8}{9}+\frac{1}{2}.\frac{2}{5}\)

=> \(S=\frac{4}{9}+\frac{1}{5}\)

=> \(S=\frac{29}{45}\)

Ta có :

\(1+\frac{1}{n\left(n+2\right)}=\frac{n\left(n+2\right)+1}{n\left(n+2\right)}=\frac{n^2+2n+1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)

Áp dụng ta được :

\(H=\frac{2.2}{1.3}\cdot\frac{3.3}{2.4}\cdot\frac{4.4}{3.5}\cdot\frac{5.5}{4.6}\cdot\frac{6.6}{5.7}\)

\(=\frac{\left(2.3.4.5.6\right)\left(2.3.4.5.6\right)}{\left(2.3.4.5\right)\left(3.4.5.6.7\right)}=\frac{6.2}{7}=\frac{12}{7}\)

Đặt \(A=\frac{1}{1.3}+\frac{1}{2.4}+...+\frac{1}{8.10}\)

\(2A=\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}\)

\(2A=1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\)

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

\(2A=\frac{9}{10}\)

\(A=\frac{9}{10}:2=\frac{9}{20}\)

=\(\frac{1}{2}\left(\frac{2}{1.3}+...+\frac{2}{8.10}\right)\)

=\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}...+\frac{1}{8}-\frac{1}{10}\right)\)

( chắc chắn có số trái dấu ở phía sau, nên còn lại như sau)

=\(\frac{1}{2}\left(1-\frac{1}{10}\right)=\frac{1}{2}.\frac{9}{10}=\frac{9}{20}\)

\(S=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}-\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+\dfrac{1}{8\cdot10}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}\right)-\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{9}\right)-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{8}{9}-\dfrac{1}{2}\cdot\dfrac{2}{5}\)

\(=\dfrac{4}{9}-\dfrac{1}{5}\)

\(=\dfrac{11}{45}\)

a) \(A=2.4+4.6+6.8+...+18.20\)

\(6A=2.4.6+4.6.\left(8-2\right)+6.8.\left(10-4\right)+...+18.20.\left(22-16\right)\)

\(6A=2.4.6+4.6.8-2.4.6+6.8.10-4.6.8+...+18.20.22-16.18.20\)

\(6A=18.20.22\)

\(A=\frac{18.20.22}{6}=\frac{7920}{6}=1320\)

d/ Đặt : A = 1.2 + 2.3 + 3.4 + ......... + 99.100

=> 3A = 1.2.(3 - 0) + 2.3.(4 - 1) + ..... + 99.100.(101 - 98)

=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ..... + 99.100.101

=> 3A = 99.100.101

=> A = 99.100.101 / 3

=> A = 333300