\(3A=\frac{3}{2.3}+\frac{3}{6.3}+\frac{1}{12.3}+\frac{3}{20.3}+\frac{3}{30.3}+\frac{3}{42.3}\)

\(3A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

\(3A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{7-6}{6.7}\)

\(3A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\)

\(3A=1-\frac{1}{7}=\frac{6}{7}\Rightarrow A=\frac{2}{7}\)

ko hiểu chi hết mà đưa ra cái sồm

ko phải tui ra đề đâu đề thi của trường chuyên vĩnh yên cấp 2 do sở ra đề

\(A=\frac{1}{6}+\frac{1}{18}+\frac{1}{36}+\frac{1}{60}+\frac{1}{90}+\frac{1}{126}\)

\(A=\frac{1}{3}\times\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)

\(A=\frac{1}{3}\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}\right)\)

\(A=\frac{1}{3}\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\right)\)

\(A=\frac{1}{3}\times\left(1-\frac{1}{7}\right)\)

\(A=\frac{1}{3}\times\frac{6}{7}=\frac{2}{7}\)

Máy tính bấm auto ra ... =))

\(A=\frac{1}{18}+\frac{1}{36}+\frac{1}{60}+...+\frac{1}{168}\)

 \(\frac{1}{3}A=\frac{1}{54}+\frac{1}{108}+...+\frac{1}{504}\)

\(\frac{1}{3}A=\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{21.24}\)

\(=\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{21}-\frac{1}{24}\)

\(=\frac{1}{6}-\frac{1}{24}\)

\(=\frac{4-1}{24}=\frac{3}{24}=\frac{1}{8}\)

=> \(A=\frac{1}{8}:\frac{1}{3}\)\(=\frac{3}{8}\)

a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)

= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + \(\dfrac{8}{1}\)

= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8

=  1 + 1 + 8

=  2 + 8

= 10

b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{10}{20}\)

=  \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (\(\dfrac{2}{2}\) + \(\dfrac{3}{3}\) + \(\dfrac{4}{4}\) + \(\dfrac{5}{5}\)+ \(\dfrac{6}{6}+\dfrac{7}{7}+\dfrac{8}{8}\) + \(\dfrac{10}{10}\))

= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (1 + 1 +1 + 1+ 1+ 1+ 1 +1)

= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x 1 x 8

= \(\dfrac{1}{2}\) + \(\)\(\dfrac{1}{2}\) x 8

= \(\dfrac{1}{2}\) + 4

= \(\dfrac{9}{2}\) 

c; \(\dfrac{1}{10}\) + \(\dfrac{4}{20}\) + \(\dfrac{9}{30}\)+\(\dfrac{16}{40}+\dfrac{25}{50}+\dfrac{36}{60}+\dfrac{49}{70}+\dfrac{64}{80}+\dfrac{81}{90}\)

= \(\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}\)

= \(\dfrac{1+2+3+4+5+6+7+8+9}{10}\)

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

= \(\dfrac{10+10+10+10+5}{10}\)

= \(\dfrac{\left(10+10+10+10\right)+5}{10}\)

= \(\dfrac{10\times4+5}{10}\)

= \(\dfrac{45}{10}\)

= \(\dfrac{9}{2}\)

a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)

  = (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + 8

= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8

= 1 + 1 + 8

= 2 + 8

= 10

b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{9}{18}\) + \(\dfrac{10}{20}\)

= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)

= \(\dfrac{1}{2}\) x 10

= 5

c; \(\dfrac{1}{10}\) + \(\dfrac{4}{20}\) + \(\dfrac{9}{30}\)+\(\dfrac{16}{40}+\dfrac{25}{50}+\dfrac{36}{60}+\dfrac{49}{70}+\dfrac{64}{80}+\dfrac{81}{90}\)

= \(\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}\)

= \(\dfrac{1+2+3+4+5+6+7+8+9}{10}\)

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

= \(\dfrac{10+10+10+10+5}{10}\)

= \(\dfrac{\left(10+10+10+10\right)+5}{10}\)

= \(\dfrac{10\times4+5}{10}\)

= \(\dfrac{45}{10}\)

= \(\dfrac{9}{2}\)

Hình như bạn viết thiếu \(\frac{1}{36}\), nếu đúng là vậy thì mình giải như sau.

Đặt A =\(\frac{1}{6}+\frac{1}{18}+\frac{1}{36}+\frac{1}{60}+\frac{1}{90}+\frac{1}{126}\)

3A = \(\frac{3}{3.2}+\frac{3}{3.6}+\frac{3}{3.12}+\frac{3}{3.20}+\frac{3}{3.30}+\frac{3}{3.42}\)

3A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

3A = \(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{7-6}{6.7}\)

3A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\)

3A = \(1-\frac{1}{7}=\frac{6}{7}\)

A = \(\frac{6}{7}:3=\frac{2}{7}\)

Gọi tổng là A:

ta có :

3A= 1/2+1/6+1/20+1/30+1/42

3A= 1/1.2+1/2.3+1/4.5+1/6.7

3A= 1-1/2+1/2-1/3+1/4-1/5+1/6-1/7

3A=1-1/3+1/4-1/5+1/6-1/7

3A= (1-1/7)+(1/3-1/4)+(1/5-1/7)

3A=6/7+-1/12+-2/35

3A=6/7-2/35+-1/12

3A=4/5-1/12

3A=43/60

A=43/180

=1/2.3+1/3.6+1/6.6+1/6.10+1/10.9+1/9.14

=1/2-1/3+1/3-1/6+1/6-1/6+1/6-1/10+1/10-1/9+1/9-1/14

=1/2-1/14

=6/14=3/7

\(\frac{1}{6}+\frac{1}{18}+\frac{1}{36}+\frac{1}{60}+\frac{1}{90}+\frac{1}{126}\)

\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot6}+\frac{1}{6\cdot6}+\frac{1}{6\cdot10}+\frac{1}{10\cdot9}+\frac{1}{9\cdot14}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}\)

\(=\frac{1}{2}-\frac{1}{14}\)

\(=\frac{3}{7}\)

cứu

a) 10; 13; 18; 26; 36; 52...

c) 0; 1; 4; 9; 16; 25...

m) 1; 4; 9; 16; 25; 36; 49; 64...

p) 1; 3; 9; 27; 81; 243...