\(pt\Leftrightarrow\left(x+1\right)\left(\frac{1}{2005}+\frac{1}{2003}-\frac{1}{2001}-\frac{1}{1999}\right)=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

\(\frac{x+1}{2005}+\frac{x+1}{2003}=\frac{x+1}{2001}+\frac{x+1}{1999}.\)

\(\Rightarrow\frac{x+1}{2005}+\frac{x+1}{2003}-\frac{x+1}{2001}-\frac{x+1}{1999}=0\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{2005}+\frac{1}{2003}-\frac{1}{2001}-\frac{1}{1999}\right)=0\)

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

\(\frac{x+1}{2005}+\frac{x+1}{2003}=\frac{x+1}{2001}+\frac{x+1}{1999}\)

\(\Rightarrow\frac{x+1}{2005}+\frac{x+1}{2003}-\frac{x+1}{2001}-\frac{x+1}{1999}=0\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{2005}+\frac{1}{2003}-\frac{1}{2001}-\frac{1}{1999}\right)=0\)

Do : \(\frac{1}{2015}+\frac{1}{2003}-\frac{1}{2001}-\frac{1}{1999}\)không thể bằng 0

\(\Rightarrow x+1=0\)

\(\Rightarrow x=0-1=-1\)

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:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}a+b=0\\b+c=0\\c+a=0\end{cases}}\)

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

\(10A=\frac{2005^{2006}+10}{2005^{2006}+1}\)

\(10B=\frac{2005^{2005}+10}{2005^{2005}+1}\)

Rồi bạn so sánh 10A và 10B là ra.

Ai thấy đúng thì ủng hộ nha !!!, sai thì góp ý cho mink nha 

Ta có

A <\(\frac{2005^{2005}+2005}{2005^{2006}+2005}=\frac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}\)=\(\frac{2005^{2004}+1}{2005^{2005}+1}\)

\(\RightarrowĐPCM\)

\(A=\frac{2005.\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)

\(B=\frac{2005.\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1+2004}{2005^{2005}+1}=1+\frac{2004}{2005^{2005}+1}\)

Vì  \(2005^{2005}+1< 2005^{2006}+1\)

\(\Rightarrow\frac{2004}{2005^{2005}+1}>\frac{2004}{2005^{2006}+1}\Rightarrow2005B>2005A\Rightarrow B>A\)