Ta có :

\(F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+..................+\dfrac{4}{2008.2010}\)

\(\Rightarrow F=2\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+.............+\dfrac{2}{2008.2010}\right)\)

\(\Rightarrow F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+..............+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)

\(\Rightarrow F=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)

\(\Rightarrow F=2.\dfrac{502}{1005}=\dfrac{1004}{1005}\)

\(F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+......+\dfrac{4}{2008.2010}\)

\(F=\dfrac{4}{2}\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+.....+\dfrac{1}{2008.2010}\right)\)

\(F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+.....+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)\(F=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)\(F=2.\dfrac{502}{1005}\)

\(F=\dfrac{1004}{1005}\)

a)\(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2008\cdot2010}\)

\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2008\cdot2010}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)

b)\(\dfrac{\dfrac{3}{41}-\dfrac{12}{47}+\dfrac{27}{53}}{\dfrac{4}{41}-\dfrac{16}{47}+\dfrac{36}{53}}=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}=\dfrac{3}{4}\)

a) gọi biểu thức đó là A

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

Dựa vào công thức trên, ta có

\(A=\dfrac{4}{2}.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+....+\dfrac{1}{2008}-\dfrac{1}{2009}\right)\)

\(A=\dfrac{4}{2}.\left(\dfrac{1}{2}-\dfrac{1}{2009}\right)\)

\(A=2.\left(\dfrac{2007}{4018}\right)=\dfrac{2007}{2009}\)

b) dễ quá bạn tự làm. (không phải mink không biết làm đâu nha)

\(A=2\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{48.50}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{50}\right)\)

\(=2\times\dfrac{12}{25}=\dfrac{24}{25}\)

\(=>A=4.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{46}-\dfrac{1}{48}+\dfrac{1}{48}-\dfrac{1}{50}\right)\)

\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=4.\left(\dfrac{25}{50}-\dfrac{1}{50}\right)=\dfrac{4.24}{50}=\dfrac{48}{25}\)

p: \(F=\dfrac{1}{3}\left(\dfrac{3}{3\cdot6}+\dfrac{3}{6\cdot9}+\dfrac{3}{9\cdot12}+...+\dfrac{3}{30\cdot33}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)

n: \(F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)

\(=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)

m: \(=\left(3-\dfrac{7}{3}+\dfrac{1}{4}\right):\left(4-\dfrac{31}{6}+\dfrac{9}{4}\right)\)

\(=\dfrac{36-28+3}{12}:\dfrac{48-62+27}{12}\)

\(=\dfrac{11}{13}\)

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

\(F=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

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

\(F=2.\frac{502}{1005}\)

\(F=\frac{1004}{1005}\)

nhinf vào là biết luật ngay bài đó bằng = \(\frac{1004}{1005}\)

kết bạn với mình nha

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

\(\Rightarrow F=2\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)

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

\(\Rightarrow F=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)

\(\Rightarrow F=2.\frac{502}{1005}\)

\(\Rightarrow F=\frac{1004}{1005}\)

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

  = 2.(1/2-1/2010)

  = 2. 502/1005

  = 1004/1005

\(b,C=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\\ =\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{30.33}\\ =\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{30}-\dfrac{1}{33}\\ =\dfrac{1}{3}-\dfrac{1}{33}\\ =\dfrac{11}{33}-\dfrac{1}{33}=\dfrac{10}{33}\)

a.F=\(\dfrac{4}{2.4}\)+\(\dfrac{4}{4.6}\)+\(\dfrac{4}{6.8}\)+...+\(\dfrac{4}{2008.2010}\)

F=\(\dfrac{2.2}{2.4}\)+\(\dfrac{2.2}{4.6}\)+\(\dfrac{2.2}{6.8}\)+...+\(\dfrac{2.2}{2008.2010}\)

F=2.(\(\dfrac{2}{2.4}\)+\(\dfrac{2}{4.6}\)+\(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{2008.2010}\))

F=2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{2008}\)-\(\dfrac{1}{2010}\))

F=2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{2010}\))

F=\(\dfrac{1004}{1005}\)

b, C=\(\dfrac{1}{18}\)+\(\dfrac{1}{54}\)+....+\(\dfrac{1}{990}\)

\(\Rightarrow\)C=\(\dfrac{1}{3.6}\)+\(\dfrac{1}{6.9}\)+...+\(\dfrac{1}{30.33}\)

=>3C=3( \(\dfrac{1}{3.6}\)+\(\dfrac{1}{6.9}\)+...+\(\dfrac{1}{30.33}\))

=>3C=\(\dfrac{3}{3.6}\)+\(\dfrac{3}{6.9}\)+....+\(\dfrac{3}{30.33}\)

=> 3C=\(\dfrac{1}{3}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{9}\)+...+\(\dfrac{1}{30}\)-\(\dfrac{1}{33}\)

=> 3C= \(\dfrac{1}{3}\)-\(\dfrac{1}{33}\)

=>3C=\(\dfrac{10}{33}\)

=> C=\(\dfrac{10}{33}\):3

=> C=\(\dfrac{10}{99}\)

a, F=4/2.4+4/4/6+4/6.8+......+4/2008.2010

=> F= 4/2.(2/2.4+2/4.6+2/6.8+......+2/2008/2010

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

=> F=4/2.( 1/2-1/2010)

=> F= 4/2. 502/1005

=> F= \(\dfrac{1004}{1005}\)

F=4/2.4+4/4.6+4/6.8+..........+4/2008.2010

F=2/2-2/4+2/4-2/6+2/6-2/8+......+2/2008-2/2010

F=2/2- 2/4+2/4-2/6+2/6-2/8+......+2/2008-2/2010

F=2/2-2/2010

=>F=2008/2010=1004/1005

\(\frac{1004}{1005}\)

Gọi F= 4/2.4+4/4.6+4/6.8+...+4/2008.2010
F/2= 2/2.4+2/4.6+...+2/2008.2010
Mà 2/2.4=1/2-1/4; 2/4.6=1/4-1/6 ....
Vậy F/2= (1/2-1/4)+(1/4-1/6)+....+(1/2008-1/2010)
F/2=1/2-1/2010=2010/4020-2/4020=2008/4...
F= 2008.2/4020=1004/1005