Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
\(S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
\(S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1006}\right)\)
\(S=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2013}\)
\(\Rightarrow\left(S-P\right)^{2013}=0^{2013}=0\).
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
Suy ra (S-P)^2013 = 0
\(2\frac{1}{117}.3\frac{1}{119}-\frac{116}{117}.5\frac{118}{119}-\frac{3}{119}=\left(3-\frac{116}{117}\right)\cdot\left(4-\frac{118}{119}\right)-5\cdot\frac{116}{117}\cdot\frac{118}{119}-\frac{3}{119}\)
mình đang ngại mình làm đến đó bạn tự phá ngoại rồi đặt nhân tử chung nha
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
\(5-\frac{x}{2010}+4-\frac{x}{2011}+3-\frac{x}{2012}=6-\frac{x}{2009}+1-\frac{x}{1007}.\)
\(\left(5+4+3\right)-x.\frac{1}{2010}-x.\frac{1}{2011}-x\frac{1}{2012}=\left(6+1\right)-x.\frac{1}{2009}-x\frac{1}{1007}\)
\(12-x.\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)=7-x.\left(\frac{1}{2009}+\frac{1}{1007}\right)\)
\(-x.\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)+x.\left(\frac{1}{2009}+\frac{1}{1007}\right)=7-12\)
\(x.\left(\frac{-1}{2010}-\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2009}+\frac{1}{1007}\right)=-5\)
\(x=\frac{-5}{\frac{-1}{2010}-\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2009}+\frac{1}{1007}}\)