a, Theo bài ra ta có:

\(M=\dfrac{2007}{1}+1+\dfrac{2006}{2}+1+.......+\dfrac{2}{2006}+1+\dfrac{1}{2007}+1-2007\)

( Ta thêm 1 vào mỗi một số hạng trong M nên phải bớt đi 2017 vì có 2017 số hạng ) ;'

\(=>M=2008+\dfrac{2008}{2}+\dfrac{2008}{3}+......+\dfrac{2008}{2007}+\dfrac{2008}{2007}-2007\)

\(=>M=\dfrac{2008}{2}+\dfrac{2008}{3}+\dfrac{2008}{4}+.....+\dfrac{2008}{2006}+\dfrac{2008}{2007}+1\)

Ta thấy xuất hiện 2008 chung nên đặt ra ngoài ta có:

\(=>M=2008\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+....+\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}\right)\)

\(=>M:N=2008\)

Câu b đợi 1 chút nha.......

b, \(M=\dfrac{1}{11.13}+\dfrac{1}{13.15}+...+\dfrac{1}{31.33}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{11.13}+\dfrac{2}{13.15}+...+\dfrac{2}{31.33}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{31}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{33}\)

\(N=\dfrac{12}{11.13.15}+\dfrac{12}{13.15.17}+...+\dfrac{12}{31.33.35}\)

\(=3\left(\dfrac{4}{11.13.15}+\dfrac{4}{13.15.17}+...+\dfrac{4}{31.33.35}\right)\)

\(=3\left(\dfrac{1}{11.13}-\dfrac{1}{13.15}+\dfrac{1}{13.15}-\dfrac{1}{15.17}+...+\dfrac{1}{31.33}-\dfrac{1}{33.35}\right)\)

\(=3\left(\dfrac{1}{11.13}-\dfrac{1}{33.35}\right)\)

\(=\dfrac{92}{5005}\)

\(\Rightarrow M:N=\dfrac{1}{33}:\dfrac{92}{5005}=\dfrac{455}{276}\)

Vậy...

g: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{19}{20}=\dfrac{1}{20}\)

h: \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot..\cdot\dfrac{100}{99}=\dfrac{100}{2}=50\)

f: \(A=1+\dfrac{1}{2^{2014}}\)

\(B=\dfrac{2^{2014}+1+1}{2^{2014}+1}=1+\dfrac{1}{2^{2014}+1}\)

mà \(2^{2014}< 2^{2014}+1\)

nên A>B

link này nè bn!

https://olm.vn/hoi-dap/detail/103540952175.html

S-P= (1 - 1/2 + 1/3 - 1/4 +...+ 1/2011 - 1/2012 + 1/2013) - ( 1/1007 + 1/1008 +...+ 1/2012 + 1/2013 )
S-P= (1- 1/2 + ... + 1/1005 - 1/1006) - 2.(1/1008 + 1/1010 + 1/1012 +...+ 1/2012)
S-P= 1+1/2+1/3+...+1/1006 - 2.( 1/2 + 1/4 + 1/6 +...+ 1/2012)
S-P= 1 + 1/2 + 1/3 +...+ 1/1006 - ( 1+ 1/2 + 1/3 +...+ 1/1006 )
S-P= 0
(S-P)^2013 = 0

\(a,\left(1-\dfrac{1}{10}\right)\times\left(1-\dfrac{1}{11}\right)\times\left(1-\dfrac{1}{12}\right)\times\left(1-\dfrac{1}{13}\right)\times\left(1-\dfrac{1}{14}\right)\times\left(1-\dfrac{1}{15}\right)=\dfrac{9}{10}\times\dfrac{10}{11}\times\dfrac{11}{12}\times\dfrac{12}{13}\times\dfrac{13}{14}\times\dfrac{14}{15}=\dfrac{9}{15}=\dfrac{3}{5}\)

\(b,\dfrac{2013\times2012-2}{2011+2011\times2013}=\dfrac{\left(2014-1\right)\times2012-2}{2011\times\left(2013+1\right)}=\dfrac{2014\times2012-2012-2}{2011\times2014}=\dfrac{2014\times2012-2014}{2011\times2014}=\dfrac{2014\times\left(2012-1\right)}{2011\times2014}=\dfrac{2011\times2014}{2011\times2014}=1\)

3/5,1

Tk mình đi mọi người mình bị âm nè!

Ai tk mình mình tk lại cho

\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)

\(=\left(1+\frac{1}{3}+......+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2012}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2012}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2012}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2013}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.......+\frac{1}{2012}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2013}\right)-\left(1+\frac{1}{2}+........+\frac{1}{1006}\right)\)

\(=\frac{1}{1007}+\frac{1}{1008}+......+\frac{1}{2013}\)

\(=P\)

\(\Leftrightarrow S-P=0\)

\(\Leftrightarrow\left(S-P\right)^{2013}=0\)

Cho mình hỏi sao lại trừ 2 lần (1/2 - 1/4 ....) thế ạ

Dạ thôi mình biết rồi ạ