TA CÓ:\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)

\(=\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{8}\)

\(=1-\frac{1}{8}=\frac{7}{8}\)

 A = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 = 7/8

= 6/12 + 2/12 + 1/12 + 3/60 + 2/60 + 8/336 + 6/336

= 9/12 + 5/60 + 14/336

= 5/6 + 1/24 = 7/8

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

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

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

A = \(9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)

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

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

A = \(\frac{81}{10}\)

\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{89}{90}=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{90}\right)\)

\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)=10-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)

\(=10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)=10-\left(1-\frac{1}{10}\right)=10-\frac{9}{10}=\frac{91}{10}\)

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)

\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)

\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)

\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

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

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

\(=9-\frac{9}{10}=\frac{81}{10}\)

=\(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{7.8}\)

=1-1/8

=7/8

D=1/1x2+1/2x3+1/3x4+1/4x5+1/5x6+1/6x7+1/7x8

=1-1/2+1/2-1/3+1/3-1/4+.........+1/7-1/8

=1-1/8=1/8

8/9 nhé

giữ lời hứa đi, k nha

8/9 nha bạn ~~~

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)

\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}...+\frac{1}{8}-\frac{1}{9}\)

\(=\frac{1}{1}-\frac{1}{9}\)

\(=1-\frac{1}{9}=\frac{8}{9}\)

A=1/2+1/6+....+1/56+1/72

A=1/1.2+1/2.3+...+1/7.8+1/8.9

A=1/1-1/2+1/2-1/3+...+1/7-1/8+1/8-1/9

A=1/1-1/9=9/9-1/9=8/9

Thanks ban nha

A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..\frac{1}{8.9}\)

A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{8}-\frac{1}{9}\)

A=1-\(\frac{1}{9}\)=\(\frac{8}{9}\)

1/6+1/12+1/20+1/30+1/42+1/56

=10/11 

nha bn

Đúng cho mk và ***** nha

thank you very much

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

\(=\frac{1}{2}-\frac{1}{8}\)

\(=\frac{3}{8}\)

tíc mình nhA

\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)

\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

\(A=\frac{1}{2}-\frac{1}{8}\)

\(A=\frac{3}{8}\)

ta có 

7/12 = 4/12 +3 /12 = 1/3 + 1/4 = 20/60 + 20/80

1/41 + 1/42 + 1/43 + ....+ 1/79 + 1/80 = ( 1/41 + 1/42 + 1/43 + ...+1/60 ) + ( 1/61 + 1/62 + 1/63 + ...+ 1/79 + 1/80 )

do 1/41 > 1/42 > 1/43 > ... > 1/59 > 1/60 

( 1/41 + 1/42 + 1/43 +...+ 1/60 ) > 1/60 + ..+ 1/60 = 20/60

và 1/61 >1/62>..1/80

( 1/61 + 1/62 + 1/63 + ...+ 1/80 ) > 1/80 +....+1/80 = 20/80

vậy 1/41 + 1/42 + 1/43 + .... + 1/79 + 1/80 > 20/60 + 20/80

1/41 + 1/42 + 1/43 + ..... + 1/79 + 1/80 > 7/12

\(\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}\)

\(=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)

\(=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)

\(=\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+\frac{9-8}{8.9}\)

\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)

\(=\frac{1}{3}-\frac{1}{9}\)

\(=\frac{3}{9}-\frac{1}{9}\)

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

1/72 = 1/9.1/8; 1/56 = 1/8.1/7; 1/42 = 1/7.1/6; 1/30 = 1/6.1/5; 1/20 = 1/5.1/4; 1/12 = 1/4.1/3

1/9 - 1/8 + 1/8 - 1/7 + 1/7 - 1/6 + 1/6 - 1/5 + 1/5 - 1/4 + 1/4 - 1/3 = 1/9 - 1/3 = -2/9

Tớ nghĩ là vậy, tớ đã tìm thấy 1 bài tương tự như vậy trên mạng rồi.

Tớ dư dấu trừ rồi. Bằng 2/9 thôi nhá, xin lỗi

A= 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90 

=1/(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) 

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5.+1/5-1/6... +1/9-1/10 

=1-1/10 

=9/10

Me too ! 

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}=x-\left(2+4+..+100\right)\)

Gọi \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)là \(A\)

         \(\left(2+4+...+100\right)\)là \(B\). Ta có :  

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)

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

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

Số số hạng của \(B\)là: \(\left(100-2\right)\div2+1=50\)

Tổng của \(B\)là: \(\left(2+100\right)\times50\div2=2550\)

\(\Rightarrow\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}=x-\left(2+4+..+100\right)=\frac{9}{10}=x-2550\)

\(\Rightarrow x=2550+\frac{9}{10}=2550+0,9=2550,9\)