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(2008 x 2009 x 2010 x 2011) x (1 + 1/2 : 3/2 - 4/3)
=(2008 x 2009 x 2010 x 2011) x (1 + 1/3 - 4/3)
=(2008 x 2009 x 2010 x 2011) x (4/3 - 4/3)
=(2008 x 2009 x 2010 x 2011) x 0
=0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x+1}{2011}+\dfrac{x+2}{2010}+\dfrac{x+3}{2009}+\dfrac{x+4}{2008}=-4\)
\(\Rightarrow\dfrac{x+1}{2011}+1+\dfrac{x+2}{2010}+1+\dfrac{x+3}{2009}+1+\dfrac{x+4}{2008}+1=0\)
\(\Rightarrow\dfrac{x+2012}{2011}+\dfrac{x+2012}{2010}+\dfrac{x+2012}{2009}+\dfrac{x+2012}{2008}=0\)
\(\Rightarrow\left(x+2012\right)\left(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}\right)=0\)
Mà \(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}\ne0\)
\(\Rightarrow x+2012=0\Rightarrow x=-2012\)
Vậy x = -2012
\(\dfrac{x+1}{2011}+\dfrac{x+2}{2010}+\dfrac{x+3}{2009}+\dfrac{x+4}{2008}=-4\\ \Leftrightarrow1+\dfrac{x+1}{2011}+1+\dfrac{x+2}{2010}+1+\dfrac{x+3}{2009}+1+\dfrac{x+4}{2008}=0\\ \Leftrightarrow\dfrac{x+2012}{2011}+\dfrac{x+2012}{2010}+\dfrac{x+2012}{2009}+\dfrac{x+2012}{2008}=0\\ \Leftrightarrow \left(x+2012\right)\left(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}\right)=0\\ \Rightarrow x+2012=0\left(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}>0\right)\\ \Rightarrow x=-2012\)
Vậy \(x=-2012\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x+1}{2011}+\dfrac{x+2}{2010}+\dfrac{x+3}{2009}+\dfrac{x+4}{2008}=-4\)
\(\Rightarrow\dfrac{x+1}{2011}+1+\dfrac{x+2}{2010}+1+\dfrac{x+3}{2009}+1+\dfrac{x+4}{2008}+1=0\)
\(\Rightarrow\dfrac{x+2012}{2011}+\dfrac{x+2012}{2010}+\dfrac{x+2012}{2009}+\dfrac{x+2012}{2008}=0\)
\(\Rightarrow\left(x+2012\right)\left(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}\right)=0\)
Mà \(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2008}\ne0\)
\(\Rightarrow x+2012=0\Rightarrow x=-2012\)
Vậy x = -2012
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\(3.\)
\(\frac{x-1}{2011}+\frac{x-2}{2010}+\frac{x-3}{2009}=\frac{x-4}{2008}\)
\(\Rightarrow\)\(\frac{x-1}{2011}-1+\frac{x-2}{2010}-1+\frac{x-3}{2009}-1-\frac{x-4}{2008}+1+2=0\)
\(\Rightarrow\)\(\frac{x-1}{2011}-\frac{2011}{2011}+\frac{x-2}{2010}-\frac{2010}{2010}+\frac{x-3}{2009}-\frac{2009}{2009}-\frac{x-4}{2008}+\frac{2008}{2008}=0\)
\(\Rightarrow\)\(\frac{x-2012}{2011}+\frac{x-2012}{2010}+\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
\(\Rightarrow\)\(x-2012\left(\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}\right)=0\)
\(\Rightarrow\)\(x=2012\)
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Theo đề bài, ta có thể viết lại như sau:
\(\left(x_1+x_2+x_3\right)+\left(x_4+x_5+x_6\right)+...+\left(x_{2008}+x_{2009}+x_{2010}\right)\)
\(=1+1+...+1\)
Vậy có số số \(1\) là:
\(2010\div3=670\)\((\)số \(1)\)
\(\Rightarrow\) Tổng trên là \(670\)
Vì tổng \(x_1+x_2+x_3+...+x_{2009}+x_{2010}=670\)nên \(x_{2011}\) là:
\(0-670=-670\)
Trả lời:\(x_{2011}=-670\)
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x1 + x2 + x3 +.....+ x2011 = 0
=> (x1 + x2) + (x3 + x4) +....+(x2009 + x2010) + x2011 = 0
=> 2 + 2 + 2 +.....+ 2 + x2011 = 0
=> 1005 . 2 + x2011 = 0
=> 2010 + x2011 = 0
=> x2011 = -2010
=> Không có giá trị nào của x thỏa mãn đề bài
![](https://rs.olm.vn/images/avt/0.png?1311)
x1 + x2 + x3 +.....+ x2011
= (x1 + x2) + (x3 + x4) +....+ ( x2009 + x2010) + x2011
= 2 + 2 + 2+.....+ 2 + x2011
= 1005 . 2 + x2011
=2010 + x2011 = 0
=> x2011 = -2010
=> Không có giá trị của x thỏa mãn đề bài
\(\frac{x+4}{2008}+\frac{x+3}{2009}=\frac{x+2}{2010}+\frac{x+1}{2011}\)
<=> \(\left(\frac{x+4}{2008}+1\right)+\left(\frac{x+3}{2009}+1\right)=\left(\frac{x+2}{2010}+1\right)+\left(\frac{x+1}{2011}+1\right)\)
<=> \(\frac{x+2012}{2008}+\frac{x+2012}{2009}=\frac{x+2012}{2010}+\frac{x+2012}{2011}\)
<=> \(\frac{x+2012}{2008}+\frac{x+2012}{2009}-\frac{x+2012}{2010}-\frac{x+2}{2011}=0\)
<=> \(\left(x+2012\right)\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)
<=> x + 2012 = 0 (Vì \(\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}\ne0\))
<=> x = -2012
Vậy x = -2012