2 + 4 + 6 + 8 + ... + 2.x = 210

=> 2.1 + 2.2 + 2.3 +2.4 + ... + 2.x = 210

=> 2.( 1 + 2 + 3 + 4 + ... +x ) = 210

=> 2. [ x.( x+ 1) /2 ] = 210

 => x. ( x + 1 ) = 210
hay x.( x + 1) = 14.(14 + 1)
Vậy x = 14

\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{18.19.20}\)

\(2A=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{20-18}{18.19.20}=\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}=\dfrac{1}{2}-\dfrac{1}{19.20}\)

\(\Rightarrow A=\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right):2\)

Đặt: \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2011.2013}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2011.2013}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2013}\right)\)

\(=\frac{1}{2}.\frac{2012}{2013}\)

\(=\frac{1006}{2013}\)

a, 6/5

b,1

1/2+1/6+1/12+1/20+...+1/380

=1/1.2+ 1/2.3+ 1/3.4+ 1/4.5+...+ 1/19.20

= 1/1-1/2+ 1/2-1/3+ 1/3-1/4+ 1/4-1/5+...+1/19-1/20

=1-1/29

=29/29-1/29

= 28/29

28/29

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{380}\)

\(=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{19\times20}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)

\(=1-\frac{1}{29}\)

\(=\frac{28}{29}\)

AI GIÚP MÌNH VỚI MÌNH ĐANG CẦN GẤP

1/3+1/6+1/12+1/24=8/24+4/24+2/24+1/24=15/24=5/8

tk cho mk nha!

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}=\frac{8}{24}+\frac{4}{24}+\frac{2}{24}+\frac{1}{24}=\frac{15}{24}=\frac{5}{8}\)

\(\frac{5}{8}\)

~Hok tốt~

Đặt A=1+1/3+1/6+1/10+ ... + 1/45

1/2.A=1/2+1/6+1/12+1/20+ ... + 1/90

1/2.A= 1/1.2+1/2.3+1/3.4+1/4.5+ ... + 1/9.10

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

1/2.A= 1-1/10

1/2.A=9/10

A=(9/10).2

A=9/5

Sai thì thôi đấy 

bài này lớp 4 mà mình không làm được nhỉ . mình học dốt dốt quá .