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

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

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

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

\(\Rightarrow x-2010=0\Rightarrow x=2010\)

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

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

\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)

\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\right)=0\Leftrightarrow x=2010\)

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

=>x-2010=0

hay x=2010

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

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

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

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

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

=> \(x-2010=0\) 

                 \(x=2010\)

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

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

=>(x-2010)x(\(\dfrac{1}{2009}\)+\(\dfrac{1}{2008}\)-\(\dfrac{1}{2007}\)-\(\dfrac{1}{2006}\))=0

=>x-2010=0 (vì \(\dfrac{1}{2009}\)+\(\dfrac{1}{2008}\)-\(\dfrac{1}{2007}\)\(\dfrac{1}{2006}\)≠0)

=>x=2010

X-1/2009 + X-2/2008 = X-3/2007 + X-4/2006

thôi nói cho nhanh nhé

bạn trừ 1 vào tất cả các giá trị VD: (X-1/2009)-1. Ta được tử chung là X-2010 cứ thế mà đặt ra làm thôi. Ko dc thì bảo tớ chỉ tiếp.

lấy mỗi phân số trừ đi 1 ta đc (x-1)/2009 -1 + (x-2)/2008 -1 = (x-3)/2007-1 + (x-4)/2006 -1

suy ra (x-2010)/2009 + (x-2010)/2008 - (x-2010)/2007 - (x-2010)/2006 = 0

đặt (x-2010) làm nhân tử chung ta được 

(x-2010).(1/2009 + 1/2008 - 1/2007 - 1/2006 ) = 0

=> x-2010 = 0 

=> x = 2010

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

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

\(\frac{x-1-2009}{2009}+\frac{x-2-2008}{2008}-\frac{x-3-2007}{2007}-\frac{x-4-2006}{2006}=0\)

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

=>(x-2010)(1/2009+1/2008-1/2007-1/2006)=0

mà 1/2009+1/2008-1/2007-1/2006 khác 0

=>x-2010=0=>x=2010

cho mìh đi rồi gửi lại đề bài qua tin nhắn cho mìh, mìh sẽ giải cho bn

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

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

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

Vì \(\frac{1}{2009}<\frac{1}{2007};\frac{1}{2008}<\frac{1}{2006}=>\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}<0\)

\(=>x-2010=0<=>x=2010\)

Sai đề rồi 

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

 Xét ta thấy \(2009\ne2008\ne2007\ne2006\)

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

Thì \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}=0\)hay \(\frac{x-1}{2009}=\frac{x-2}{2008}=\frac{x-3}{2007}=\frac{x-4}{2006}=1\)

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

Không thể bằng 0 được

Ta có \(\frac{x-1}{2009}=\frac{x-2}{2008}=\frac{x-3}{2007}=\frac{x-4}{2006}=1\) Nên \(x-1=2009;x-2=2008;x-3=2007;x-4=2006\)

Suy ra \(x=2010\)P/S: Sở dĩ \(\frac{x-1}{2009}=\frac{x-2}{2008}=\frac{x-3}{2007}=\frac{x-4}{2006}=1\)

được là bởi vì \(2009=2010-1\)và \(2008=2010-2\)và \(2007=2010-3\)và \(2006=2010-4\)