B= 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + ...+ 2009 - 2010 - 2011 + 2012

B=( 1 - 2 - 3 + 4 )+( 5 - 6 - 7 + 8) + ...+ (2009 - 2010 - 2011 + 2012)

B = 0 + 0 + ... + 0

B = 0

\(A=\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...+\frac{4}{2010\cdot2012}\)

\(\Rightarrow\frac{2}{4}A=\frac{2}{4}\left(\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+.....+\frac{4}{2010\cdot2012}\right)\)

\(\Rightarrow\frac{2}{4}A=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2010\cdot2012}\)

\(\Rightarrow\frac{2}{4}A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2010}-\frac{1}{2012}\)

\(\Rightarrow\frac{2}{4}A=\frac{1}{2}-\frac{1}{2012}\Rightarrow\frac{2}{4}A=\frac{1005}{2012}\)

\(\Rightarrow A=\frac{1005}{2012}:\frac{2}{4}\Rightarrow A=\frac{1005}{1006}\)

May cai gach la phan nhe

\(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2010.2012}\)

\(A=2\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2010.2012}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2010}-\frac{1}{2012}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{2012}\right)=2\left(\frac{1006}{2012}-\frac{1}{2012}\right)=2.\frac{1005}{2012}\)

\(A=\frac{2.1005}{2012}=\frac{1.1005}{1006}=\frac{1005}{1006}\)

Số số hạng tính từ 2 đến 2012: (2012 - 2) : 2 + 1 = 1006

Ta nhóm: (2 - 4) + (6 - 8) + ....+ (2010 - 2012)  

Mỗi nhóm = - 2 => có 503 nhóm

Vậy: 2 - 4 + 6 - 8 + 10+...+ 2010 - 2012 + 2014 = (-2) . 503 + 2014 = 1008

2012-0:2+1=1007

[2012+0]x1007:2=1013042

a, 1+2-4+6-8+10-12+...+2010-2012+2013

=1+(4-2)+(6-8)+(12-10)+...+(2012-2010)+2013

=1+2+2+2+...+2+2013 ( 503 số 2 )

=1+503x2+2013

=2014+1006

=3020

  1+2-4+6-8+10-12+...2010-2012+2013

= ( 1 + 2013 ) +  ( 2 - 4 ) + ( 6 - 8 ) + ( 10 - 12 ) + ....... + ( 2010 - 2012 ) 

= 2004 + (-2) + (-2) + (-2) + (-2) +........ + (-2)

                    ( có 503 số - 2 ) 

=    2004 + (-2) . 503 

=   2004 + (-1006)

=      998