\(\frac{1}{1}:2+\frac{1}{2}:3+\frac{1}{3}:4+...+\frac{1}{2009}:2010+\frac{1}{2010}:2011\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}+\frac{1}{2010.2011}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}+\frac{1}{2010}-\frac{1}{2011}\)

\(=1-\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{2009}-\frac{1}{2009}\right)+\left(\frac{1}{2010}-\frac{1}{2010}\right)-\frac{1}{2011}\)

\(=1-\frac{1}{2011}=\frac{2010}{2011}\)

~ Hok tốt ~

\(\frac{1}{1}:2+\frac{1}{2}:3+\frac{1}{3}:4+...+\frac{1}{2009}:2010+\frac{1}{2010}:2011\)

\(=\frac{1}{1}:\frac{2}{1}+\frac{1}{2}:\frac{3}{1}+\frac{1}{3}:\frac{4}{1}+...+\frac{1}{2009}:\frac{2010}{1}+\frac{1}{2010}:\frac{2011}{1}\)

\(=\frac{1}{1}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+...+\frac{1}{2009}\cdot\frac{1}{2010}+\frac{1}{2010}\cdot\frac{1}{2011}\)

\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2010}+\frac{1}{2010\cdot2011}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\)

\(=1-\frac{1}{2011}=\frac{2010}{2011}\)

Dấu " . " là dấu nhân nhé

a. 1⋅2⋅3+2⋅4⋅6+3⋅6⋅9+4⋅8⋅12

= 6+2⋅4⋅6+3⋅6⋅9+4⋅8⋅12

= 6+48+3⋅6⋅9+4⋅8⋅12

= 6+48+162+4⋅8⋅12

= 6+48+162+384

= 600

b . Ta có \(A=\frac{2010+2011}{2011+2012}=\frac{2010}{2011+2012}+\frac{2011}{2011+2012}.\)

Ta có : \(\frac{2010}{2011+2012}< \frac{2010}{2011}\) và \(\frac{2011}{2011+2012}< \frac{2011}{2012}\)

=> \(\frac{2010+2011}{2011+2012}< \frac{2010}{2011}+\frac{2011}{2012}\)

=> A < B

=211 nếu đúng thì k ha

= ((5,2 x 5 + 0,25 ) x 4 + 1/2 ) x 2 

= ( 26,25 x 4 + 0,5) x 2

= 105,5 x 2

= 211

Hình như đề bài phải là : Tính tổng : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}+\frac{1}{2010.2011}\)

Nếu thế giải như sau : \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}+\frac{1}{2010}-\frac{1}{2011}\)

\(=1-\frac{1}{2011}=\frac{2010}{2011}.\)Vậy tổng đó là 2010/2011.

Ta có :\(\frac{1}{1}:2+\frac{1}{2}:3+...+\frac{1}{2010}:2011\)

= \(\frac{1}{1}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{3}+...+\frac{1}{2010}\times\frac{1}{2011}\)

= \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2010\times2011}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)

= \(1-\frac{1}{2011}\)

= \(\frac{2010}{2011}\)

B=2/1.3 + 2/3.5 + 2/5.7 +...+ 2/299.301

B=1-1/3+1/3-1/5+1/5-1/7+...+1/299-1/301=1-1/301=300/301

\(Ta có: \frac{2}{3}=\frac{1}{1}-\frac{1}{3}\);

\(\frac{2}{15}=\frac{1}{3}-\frac{1}{5}\);

\(\frac{2}{35}=\frac{1}{5}-\frac{1}{7}\) ; ... ; \(\frac{2}{89999}=\frac{1}{299}-\frac{1}{301}\).

=> B= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{299}-\frac{1}{301}\)

=> B=\(\frac{1}{1}-\frac{1}{301}\)

=> B=\(\frac{300}{301}\)