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Ta có:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}a+b=0\\b+c=0\\c+a=0\end{cases}}\)
Với \(a+b=0\)
Thì \(\hept{\begin{cases}\frac{1}{a^{2005}}+\frac{1}{b^{2005}}+\frac{1}{c^{2005}}=\frac{1}{c^{2005}}\\\frac{1}{a^{2005}+b^{2005}+c^{2005}}=\frac{1}{c^{2005}}\end{cases}}\)
Tương tự cho 2 trường hợp còn lại ta có ĐPCM
Với \(a,b,c\ne0\); \(a+b+c\ne0\) , ta có:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\left(a+b+c\right)\left(ab+bc+ca\right)=abc\)
\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca\right)+c\left(ab+bc+ca\right)=abc\)
\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca\right)+abc+bc^2+c^2a=abc\)
\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca\right)+bc^2+c^2a=0\)
\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca\right)+c^2\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca+c^2\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left[b\left(a+c\right)+c\left(a+c\right)\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b=0\\b+c=0\\c+a=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)
Không mất tính tổng quát, ta lấy \(a=-b\), ta có:
\(\frac{1}{a^{2005}}+\frac{1}{b^{2005}}+\frac{1}{c^{2005}}=\frac{1}{\left(-b\right)^{2005}}+\frac{1}{b^{2005}}+\frac{1}{c^{2005}}\)
\(=\frac{-1}{b^{2005}}+\frac{1}{b^{2005}}+\frac{1}{c^{2005}}=\frac{1}{c^{2005}}\) (1)
Ta có:\(\frac{1}{a^{2005}+b^{2005}+c^{2005}}=\frac{1}{\left(-b\right)^{2005}+b^{2005}+c^{2005}}\)
\(=\frac{1}{-b^{2005}+b^{2005}+c^{2005}}=\frac{1}{c^{2005}}\) (2)
Từ (1), (2), suy ra \(\frac{1}{a^{2005}}+\frac{1}{b^{2005}}+\frac{1}{c^{2005}}=\frac{1}{a^{2005}+b^{2005}+c^{2005}}\)
Cái chỗ không mất tính tổng quát đấy, là do a, b, c bình đẳng nhau.
\(pt\Leftrightarrow\left(x+1\right)\left(\frac{1}{2005}+\frac{1}{2003}-\frac{1}{2001}-\frac{1}{1999}\right)=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
\(\frac{x+1}{2005}+\frac{x+1}{2003}=\frac{x+1}{2001}+\frac{x+1}{1999}.\)
\(\Rightarrow\frac{x+1}{2005}+\frac{x+1}{2003}-\frac{x+1}{2001}-\frac{x+1}{1999}=0\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{2005}+\frac{1}{2003}-\frac{1}{2001}-\frac{1}{1999}\right)=0\)
Mà \(\frac{1}{2005}+\frac{1}{2003}-\frac{1}{2001}-\frac{1}{1999}#0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
Vậy nghiệm của pt là x = -1
Ta có: \(2-x+2005=1-x+2006=-x+2007\)
\(\frac{2-x}{2005}-1=\frac{1-x}{2006}-\frac{x}{2007}\)
\(\Leftrightarrow\frac{2-x}{2005}+1-2=\frac{1-x}{2006}+1+\left(\frac{-x}{2007}+1\right)-2\)
\(\Leftrightarrow\frac{2007-x}{2005}=\frac{2007-x}{2006}+\frac{2007-x}{2007}\)
\(\Leftrightarrow\left(2007-x\right)\left(\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\right)=0\)
\(\Rightarrow2007-x=0\)
\(\Rightarrow x=2007\)
\(\frac{2-x}{2005}-1=\frac{1-x}{2006}-\frac{x}{2007}\)
\(\Leftrightarrow\frac{2-x}{2005}-\frac{1-x}{2006}+\frac{x}{2007}-1=0\)
\(\Leftrightarrow\frac{2-x}{2005}+1-\frac{1-x}{2006}-1+\frac{x}{2007}-1=0\)
\(\Leftrightarrow\left(\frac{2-x}{2005}+1\right)-\left(\frac{1-x}{2006}+1\right)-\left(1-\frac{x}{2007}\right)=0\)
\(\Leftrightarrow\frac{2-x+2005}{2005}-\frac{1-x+2006}{2006}-\frac{2007-x}{2007}=0\)
\(\Leftrightarrow\frac{2007-x}{2005}-\frac{2007-x}{2006}-\frac{2007-x}{2007}=0\)
\(\Leftrightarrow\left(2007-x\right)\left(\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\right)=0\)
\(\Leftrightarrow2007-x=0\) < Vì \(\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\ne0\)>
\(\Leftrightarrow x=2007\)
VẬY \(x=2007\)
Ta có : \(\frac{x^2-2008}{2007}+\frac{x^2-2007}{2006}+\frac{x^2-2006}{2005}=\frac{x^2-2005}{2004}+\frac{x^2-2004}{2003}+\frac{x^2-2003}{2002}\)
=> \(\frac{x^2-2008}{2007}+1+\frac{x^2-2007}{2006}+1+\frac{x^2-2006}{2005}+1=\frac{x^2-2005}{2004}+1+\frac{x^2-2004}{2003}+1+\frac{x^2-2003}{2002}+1\)
=> \(\frac{x^2-2008}{2007}+\frac{2007}{2007}+\frac{x^2-2007}{2006}+\frac{2006}{2006}+\frac{x^2-2006}{2005}+\frac{2005}{2005}=\frac{x^2-2005}{2004}+\frac{2004}{2004}+\frac{x^2-2004}{2003}+\frac{2003}{2003}+\frac{x^2-2003}{2002}+\frac{2002}{2002}\)
=> \(\frac{x^2-1}{2007}+\frac{x^2-1}{2006}+\frac{x^2-1}{2005}=\frac{x^2-1}{2004}+\frac{x^2-1}{2003}+\frac{x^2-1}{2002}\)
=> \(\frac{x^2-1}{2007}+\frac{x^2-1}{2006}+\frac{x^2-1}{2005}-\frac{x^2-1}{2004}-\frac{x^2-1}{2003}-\frac{x^2-1}{2002}=0\)
=> \(\left(x^2-1\right)\left(\frac{1}{2007}+\frac{1}{2006}+\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\right)=0\)
=> \(x^2-1=0\)
=> \(x^2=1\)
=> \(x=\pm1\)
Vậy phương trình có 2 nghiệm là x = 1, x = -1 .