=1/2 + (1/2*3+1/3*4) + (1/4*5+1/5*6) + (1/6*7+1/7*8) + (1/8*9+1/9*10)
=1/2 + 1/2*3.(1+1/2) + 1/2*5.(1/2+1/3) + 1/2*7.(1/3+1/4) + 1/2*9.(1/4+1/5)
=1/2 + 1/2*3.(3/2) + 1/2*5.(5/6) + 1/2*7.(7/12) + 1/2*9.(9/20)
=1/2 + 1/4 + 1/12 + 1/24 + 1/40
=9/10

6 = 2 x 2 + 2 = 2(2+1)
12 = 3 x 3 + 3 = 3(3+1)
20 = 4 x 4 + 4 = 4(4+1)
...
90 = 9 x 9 + 9 = 9(9+1)

Ta có đồng nhất thức sau 
1/[n(n+1)] = 1/n - 1/(n+1)

Vậy
     1/6 = 1/2 - 1/3
     1/12 = 1/3 - 1/4
     1/20 = 1/4 - 1/5
....
     1/90 = 1/9 - 1/110

S =  1/2 -1/3+1/3-1/4+..............+1/9 -1/10

Tổng là 1/2 + 1/2 - 1/10 = 1 - 1/10 = 9/10
 

\(S=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)

\(\Rightarrow S=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

       \(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)(áp dụng quy tắc dấu ngoặc )

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

      \(S=\frac{1}{2}-\frac{1}{10}\) 

      \(S=\frac{2}{5}\)

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{89}{90}\)

= \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{90}\right)\)

= \(\left(1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)8 số hạng 1

= \(\left(1.8\right)-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)

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

= \(8-\left(1-\frac{1}{10}\right)\)

= \(8-\frac{9}{10}\)

= \(\frac{71}{10}\)

A=\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+......+\(\frac{1}{9.10}\)

A=\(\frac{1}{1}\)-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+........+\(\frac{1}{9}\)-\(\frac{1}{10}\)

A=1-\(\frac{1}{10}\)=9/10

Mơn thím nhìu

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

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

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

Giải hộ mik nha mấy bạn

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{90}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{9}-\frac{1}{10}\)

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

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

=1/1x2+1/2x3+1/3x4+.....+1/9x10

=1/1-1/2+1/2-1/3+1/3-1/4+.........-1/10

=1/1-1/10

=1/9

1/2+1/6+....+1/90

=1/1.2 +1/2.3 +1/3.4+.....+1/9.10

=1-1/2+1/2-1/3+1/3-1/4+...+1/9-1/10

=1-1/10=9/10

k 3 cái sao đc

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

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

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

\(=>S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{48.49}+\frac{1}{49.50}\)

\(=>S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}=\frac{1}{1}-\frac{1}{50}=\frac{49}{50}\)

vậy S=49/50

49/50 nha bn

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{90}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{9}-\frac{1}{10}\)

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

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

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{30}+...+\frac{1}{90}\)

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)

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

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

\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\)

\(S=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)

\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

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

\(S=\frac{9}{10}\)

\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\)

\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

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

\(S=\frac{9}{10}\)

Ta có:

\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\)

\(\Rightarrow S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(\Rightarrow S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

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

\(\Rightarrow S=\frac{9}{10}\)