\(\frac{x+3}{2007}-\frac{x+3}{2008}=\frac{x+3}{2010}-\frac{x+3}{2009}\)

\(\Leftrightarrow\)\(\frac{x+3}{2007}-\frac{x+3}{2008}-\frac{x+3}{2010}+\frac{x+3}{2009}=0\)

\(\Leftrightarrow\) \(\left(x+3\right).\left(\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\right)=0\)

\(\Leftrightarrow\) \(x+3=0\) ( Vì \(\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\ne0\) )

\(\Leftrightarrow\) \(x=-3\)

Vậy x = -3

\(\frac{x+3}{2007}-\frac{x+3}{2008}=\frac{x+3}{2010}-\frac{x+3}{2009}\)

\(\Rightarrow\frac{x+3}{2007}-\frac{x+3}{2008}-\frac{x+3}{2010}+\frac{x+3}{2009}=0\)

\(\Rightarrow\left(x+3\right)\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\right)=0\)

\(\Rightarrow x+3=0\Leftrightarrow x=-3\)

\(\frac{x+3}{2007}-\frac{x+3}{2008}=\frac{x+3}{2010}-\frac{x+3}{2009}\)

\(\Rightarrow\frac{x+3}{2007}-\frac{x+3}{2008}-\frac{x+3}{2010}+\frac{x+3}{2009}=0\)

\(\Rightarrow\left(x+3\right)\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x+3=0\\\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\ne0\end{cases}\hept{\begin{cases}x=-3\\\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\ne0\end{cases}}}\)

\(\Rightarrow x=-3\)

\(\frac{x+3}{2007}-\frac{x+3}{2008}=\frac{x+3}{2010}-\frac{x+3}{2009}\)

\(\Leftrightarrow\)\(\frac{x+3}{2007}-\frac{x+3}{2008}-\frac{x+3}{2010}+\frac{x+3}{2009}=0\)

\(\Leftrightarrow\)\(\left(x+3\right)\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\right)=0\)

Vì \(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\ne0\)

Nên \(x+3=0\)

\(\Leftrightarrow\)\(x=-3\)

Vậy \(x=-3\)

Chúc bạn học tốt ~ 

\(\left(\frac{x+4}{2007}+1\right)+\left(\frac{x+3}{2008}+1\right)=\left(\frac{x+2}{2009}+1\right)+\left(\frac{x+1}{2010}+1\right)\)

\(\left(\frac{x+2011}{2007}\right)+\left(\frac{x+2011}{2008}\right)=\left(\frac{x+2011}{2009}\right)+\left(\frac{x+2011}{2010}\right)\)
\(\frac{x+2011}{2007}+\frac{x+2011}{2008}-\frac{x+2011}{2009}-\frac{x+2011}{2010}=0\)

\(\left(x+2011\right).\left(\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)=0\)

Vì \(\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\)khác 0 (các số hạng ko bằng nhau)

\(\Leftrightarrow\)\(x+2011=0\)

\(\Rightarrow x=0-2011\)

\(\Rightarrow x=-2011\)

2010 nha

\(\Rightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)=\left(\frac{x+3}{2007}+1\right)+\left(\frac{x+4}{2006}+1\right)\)

\(\Rightarrow\left(\frac{x+1}{2009}+\frac{2009}{2009}\right)+\left(\frac{x+2}{2008}+\frac{2008}{2008}\right)=\left(\frac{x+3}{2007}+\frac{2007}{2007}\right)+\left(\frac{x+4}{2006}\frac{2006}{2006}\right)\)

\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}=\frac{x+2010}{2007}+\frac{x+2010}{2006}\)

\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}-\frac{x+2010}{2007}-\frac{x+2010}{2006}=0\)

\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)

Vì \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\)

=>x+2010=0

=>x=-2010

Vậy x = -2010

chuanhieu cho lam

Trừ 1 đi ở mỗi phân số, ta có:

\(\frac{x-1}{2009}-1+\frac{x-2}{2008}-1=\frac{x-3}{2007}-1+\frac{x-4}{2006}-1\)

\(\Rightarrow\frac{x-1}{2009}-\frac{2009}{2009}+\frac{x-2}{2008}-\frac{2008}{2008}=\frac{x-3}{2007}-\frac{2007}{2007}+\frac{x-4}{2006}-\frac{2006}{2006}\)

\(\Rightarrow\frac{x-1-2009}{2009}+\frac{x-2-2008}{2008}=\frac{x-3-2007}{2007}+\frac{x-4-2006}{2006}\)

\(\Rightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}=\frac{x-2010}{2007}+\frac{x-2010}{2006}\)

\(\Rightarrow\left[x-2010\right]\left[\frac{1}{2009}+\frac{1}{2008}\right]=\left[x-2010\right]\left[\frac{1}{2007}+\frac{1}{2006}\right]\)

Sẽ có hai trường hợp 

TH1: Cả hai vế đều bằng 0

Ta có: \(\hept{\begin{cases}\frac{1}{2009}+\frac{1}{2008}\ne0\\\frac{1}{2007}+\frac{1}{2006}\ne0\end{cases}}\Rightarrow x-2010=0\Rightarrow x=2010\)

TH2: Cả hai vế khác 0

Ta bỏ đi x - 2010 vì cả hai bên đều có

\(\Rightarrow\frac{1}{2009}+\frac{1}{2008}=\frac{1}{2007}+\frac{1}{2006}\)Vô lí

Vậy x = 2010

Ta có :

\(\frac{x+1}{2012}+\frac{x+2}{2011}+\frac{x+3}{2010}=\frac{x+4}{2009}+\frac{x+5}{2008}+\frac{x+6}{2007}\)

\(\left(\frac{x+1}{2012}+1\right)+\left(\frac{x+2}{2011}+1\right)+\left(\frac{x+3}{2010}+1\right)=\left(\frac{x+4}{2009}+1\right)+\left(\frac{x+5}{2008}+1\right)+\left(\frac{x+6}{2007}+1\right)\)

\(\Leftrightarrow\)\(\frac{x+2013}{2012}+\frac{x+2013}{2011}+\frac{x+2013}{2010}=\frac{x+2013}{2009}+\frac{x+2013}{2008}+\frac{x+2013}{2007}\)

\(\Leftrightarrow\)\(\left(x+2013\right).\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}\right)=\left(x+2013\right).\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\right)\)

\(\Leftrightarrow\)\(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}=\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\)\(\left(1\right)\)

Mà \(\frac{1}{2012}< \frac{1}{2009}\)\(;\)\(\frac{1}{2011}< \frac{1}{2008}\)\(;\)\(\frac{1}{2010}< \frac{1}{2007}\)

\(\Rightarrow\)\(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}< \frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\)\(\left(2\right)\)

Từ \(\left(1\right)\)và \(\left(2\right)\)suy ra không có giá trị nào của \(x\)thoả mãn đề bài 

Vậy không có gía trị nào của \(x\)hay \(x\in\left\{\varnothing\right\}\)