Ta có phương trình :
\(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+....+\frac{x-2012}{1}=2012\)
Ta thấy phương trình đã cho tương ứng với phương trình :
\(\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+...+\left(\frac{x-2012}{1}-1\right)+2012=2012\)
\(\Rightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)
\(\Rightarrow\left(x-2013\right).\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+....+1\right)=0\)
Mặt khác \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+1\ne0\)
Do đó \(\Rightarrow x-2013=0\Rightarrow x=2013\)
Do vậy \(x=2013\)thoả mãn phương trình ban đầu
\(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2000}+.....+\frac{x-2012}{1}=2012\)
\(\Leftrightarrow\frac{x-1}{2012}+\frac{x-2}{2011}+........+\frac{x-2012}{1}-2012=0\)
\(\Leftrightarrow\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+......+\left(\frac{x-2012}{1}-1\right)=0\)
\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+......+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+.....+1\right)=0\)
Mà \(\frac{1}{2012}+\frac{1}{2011}+....+1\ne0\)
Vậy ...
\(\Leftrightarrow x=2013\)
\(\Leftrightarrow x-2013=0\)
pt <=> (x-1/2012 - 1) + (x-2/2011 - 1) + ...... + (x-2012/1 - 1) = 0
<=> x-2013/2012 + x-2013/2011 + ...... + x-2013/1 = 0
<=> (x-2013).(1/2012 + 1/2011 + ..... + 1) = 0
<=> x-2013 = 0 ( vì 1/2012 + 1/2011 + ..... + 1 > 0 )
<=> x=2013
Vậy pt có tập nghiệm S = {2013}
Tk mk nha
\(\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+....+\left(\frac{x-2012}{1}-1\right)=0\)
\(\Rightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)
\(\Rightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+1\right)=0\)
\(\Rightarrow x=2013\)