Ta gọi biểu thức đó là A

Ta có công thức        \(\frac{a}{b.c}=\frac{a}{c-b}.\left(\frac{1}{b}-\frac{1}{c}\right)\)

Dựa vào công thức ta có  

\(\frac{4}{2.4}=2.\left(\frac{1}{2}-\frac{1}{4}\right)\)

\(\frac{4}{4.6}=2.\left(\frac{1}{4}-\frac{1}{6}\right)\)

\(....................\)

\(\frac{4}{18.20}=2.\left(\frac{1}{18}-\frac{1}{20}\right)\)

\(\Rightarrow\)\(A=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{18}-\frac{1}{20}\right)\)

\(\Rightarrow\)\(A=2.\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(\Rightarrow\)\(A=2.\left(\frac{9}{20}\right)=\frac{18}{20}\)

Ai thấy đúng thì ủng hộ nha !!!

sai rồi kết quả phải bằng 9/10 chứ 

\(\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...+\frac{4}{18\cdot20}\)

\(=2\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{18\cdot20}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{18}-\frac{1}{20}\right)\)

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

\(=2\left(\frac{9}{20}\right)\)\(=\frac{18}{20}=\frac{9}{10}\)

ax4=1/2-1/4+1/4-1/6+1/6+1/8+.......+1/16-1/18+1/18+1/20

ax4  =1/2-1/20

 ax4 =9/20

   a=9/20:4

   a=9/80

THANKS

K = đề bài

   = 2 . ( 2/2.4 + 2/4.6 + 2/6.8 + . . . + 2/2008.2010 )

   = 2 . ( 1 - 1/4 + 1/4 - 1/6 + 1/8 - 1/8 + . . . + 2/2008 - 2/2010 )

   = 2 . ( 1 - 2/2010 )

   = ( phần còn lại bạn tự tính nha )

k cho mình đó, bài này mình làm rồi nên đúng 100% lun, sorry nha mình ngại viết nhiều

a) \(K=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)

\(\Leftrightarrow K=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\right)\)

\(\Leftrightarrow K=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(\Leftrightarrow K=2.\left(1-\frac{1}{2010}\right)\)

\(\Leftrightarrow K=2.\frac{2009}{2010}=\frac{2009}{1005}\)

b)  F=1/18 + 1/54 + 1/108 +...+ 1/990

=> \(F=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)

\(\Leftrightarrow F=3.\left(\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\right)\)

\(\Leftrightarrow F=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\)

\(\Leftrightarrow F=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\)

\(\Leftrightarrow F=1-\frac{1}{33}=\frac{32}{33}\)

\(17.8+51.4=34.4+51.4=4\left(51+34\right)=4.84=336\) \(2.2.3.5.19=\left(2.5\right).\left(3.19\right).2=10.2.57=570.2=1140\) \(54.275+825.15+275=54.275+45.275+275=275\left(54+45+1\right)=100.275=27500\) \(\frac{167.198+98}{198.168-100}=\frac{167.198+98}{198.167+198-100}=\frac{167.198+98}{167.198+98}=1\)

\(\frac{1}{n}-\frac{1}{n+k}=\frac{k}{n\left(n+k\right)}\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{2020}=1-\frac{1}{2020}=\frac{2019}{2020}\)

a) 17 x 8 + 51 x 4

= 17 x 4 x 2 + 17 x 3 x 4

= 17 x 4 x ( 2 + 3 )

= 14 x 4 x 5

= 14 x 20

= 280

b) 2 x 2 x 3 x 5 x 19

= ( 2 x 5 ) x ( 3 x 19 ) x 2

= 10 x 57 x 2

= 570 x 2

= 1140

c) 54 x 275 + 825 x 15 + 275

= 54 x 275 + 275 x 3 x 15 + 275 x 1

= 54 x 275 + 275 x 45 + 275 x 1

= 275 x ( 54 + 45 + 1 )

= 275 x 100

= 27500

d) 100 - 99 + 98 - 97 + 96 - 95 + 94 - 93 + ... + 4 - 3 + 2

= (100 - 99) + (98 - 97) + (96 - 95) + (94 - 93) + ... + (4 - 3) + 2

= (1 + 1 + ... + 1) + 2

( 49 số 1 )

= 49 + 2

= 51

k) 1,5 + 2,5 + 3,5 + 4,5 + 5,5 + 6,5 + 7,5 + 8,5

= ( 1,5 + 8,5 ) + ( 2,5 + 7,5 ) + ( 3,5 + 6,5 ) + ( 4,5 + 5,5 )

= 10 + 10 + 10 + 10

= 40

m) Đề bn xem ở trên

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

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

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

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

Đặt:A =  \(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+.....+\frac{4}{2008.2010}\)

=> A = 2.(\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+.....+\frac{2}{2008.2010}\)

=> A = 2.(\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{2008}-\frac{1}{2010}\)

=> A = 2.(\(\frac{1}{2}-\frac{1}{2010}\))

=> A = 2.\(\frac{502}{1005}\)

=> A = \(\frac{1004}{1005}\)

đặt A= \(\frac{4}{2.4}+\frac{4}{4.6}+...+\frac{4}{2008.2010}\) 

=> 1/2.A=\(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\) 

= \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\)

=\(\frac{1}{2}-\frac{1}{2010}\)

=\(\frac{502}{1005}\)

Vậy biểu thức cần tìm có giá trị là \(\frac{502}{1005}\)

nhầm nha

=> A= \(\frac{502}{1005}:\frac{1}{2}\)=\(\frac{1004}{1005}\)

\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)=\frac{2.2004}{2010}=\frac{2004}{1005}\)

\(=\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+...+\frac{2}{1004\cdot1005}\)

\(=2\cdot\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{1004\cdot1005}\right)\)

\(=2\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1004}-\frac{1}{1005}\right)\)

\(=2\cdot\left(1-\frac{1}{1005}\right)=2\cdot\frac{1004}{1005}=\frac{2008}{1005}\)