giúp tui viết cả lời giải nữa tui sắp nộp zùi ai trả lời đúng tui k 3 cái cho

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 

Lời giải:

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

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

\(=1+\underbrace{(-2)+(-2)+....+(-2)}_{503}\)

\(=1+(-2).503=-1005\)

\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

\(A=7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{69.70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{69}-\frac{1}{70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)\)

\(A=7\cdot\frac{3}{35}=\frac{21}{35}\)

\(A=\frac{7}{10\cdot11}+\frac{7}{11\cdot12}+\frac{7}{12\cdot13}+...+\frac{7}{69\cdot70}\)

\(A=7\left(\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+\frac{1}{12\cdot13}+...+\frac{1}{69\cdot70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)=7\cdot\frac{3}{35}=\frac{3}{5}\)

\(B=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\)

\(B=\frac{1}{2}\left(\frac{2}{25\cdot27}+\frac{2}{27\cdot29}+\frac{2}{29\cdot31}+...+\frac{2}{73\cdot75}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)=\frac{1}{2}\cdot\frac{2}{75}=\frac{1}{75}\)

\(C=\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...+\frac{4}{2008\cdot2010}\)

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

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

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

\(B=\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)

\(2B=\frac{2}{25.27}+\frac{2}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)

\(2B=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\)

\(2B=\frac{1}{25}-\frac{1}{75}=\frac{2}{75}\)

2012-0:2+1=1007

[2012+0]x1007:2=1013042

Bạn có viết lộn đề ko ?

No , I do not

Ta có : 

\(2012^2.A=2012^{12}-2012^{10}=B\)

Vì \(2012^2.A=B\) nên \(A< B\)

Vậy \(A< B\)

Chúc bạn học tốt ~