\(\frac{1}{2}.4+\frac{1}{4}.6+\frac{1}{6}.8+\frac{1}{8}.10\)

=\(\frac{73}{12}\)

k mình nha

18:

a: \(S=3\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{98\cdot100}\right)\)

=3*(1/2-1/4+1/4-1/6+...+1/98-1/100)

=3*49/100=147/100

b: Để A là số nguyên thì n-1 thuộc Ư(2)

=>n-1 thuộc {1;-1;2;-2}

=>n thuộc {2;0;3;-1}

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

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

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

=2.(1/2-1/2010)

=2.502/1005

=1004/1005

Vậy A=1004/1005

100% giải đúng đầu tiên:

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

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

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

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

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

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

                       \(=1-\frac{1}{1005}=\frac{1004}{1005}\)

a: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{47}-\dfrac{4}{47}+\dfrac{9}{53}\right)}=\dfrac{3}{4}\)

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

\(=2\cdot\dfrac{1004}{2010}=\dfrac{2008}{2010}=\dfrac{1004}{1005}\)

c: \(S=\dfrac{1}{3\cdot6}+\dfrac{1}{6\cdot9}+...+\dfrac{1}{30\cdot33}\)

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

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

=> K : 2 = \(\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{2008.2010}\)

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

            =\(\frac{1}{2}-\frac{1}{2010}=\frac{502}{1005}\)

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

Vậy \(K=\frac{1004}{1005}\)

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

  = 2.(1/2-1/2010)

  = 2. 502/1005

  = 1004/1005

Ta có: \(F=\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2008\cdot2010}\)

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

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

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

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

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

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

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

\(F=1-\dfrac{1}{1005}=\dfrac{1004}{1005}\)

=2(2/2.4+2/4.6+......+22/008.2010)

=2(12−12010)

:D thầy ra đề cương khó quá nên các bạn giúp mình với!!!!!

S = \(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2016.2018}\)

S = \(2.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2016.2018}\right)\)

S = \(2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2016}-\frac{1}{2018}\right)\)

S = \(2.\left(\frac{1}{2}-\frac{1}{2018}\right)\)

S = \(2.\frac{504}{1009}\)= \(\frac{1008}{1009}\)

Vậy S = \(\frac{1008}{1009}\).

~~~

Nếu có sai sót gì thì giúp đỡ tớ nha :3

#Sunrise

Cảm ơn Lê Cao Mai Anh nhìu lắm