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

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

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

Ta có:

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

=> (x-1/2009 - 1)+(x-2/2008 - 1) = (x-3/2007 - 1) + (x-4/2006 -1 )

=> x-2010/2009 + x-2010/2008 = x-2010/2007 + x-2010/2006

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

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

Vì (1/2009+1/2008-1/2007-1/2006) KHÁC 0

=>x-2010=0

=>x=2010

Cảm ơn bạn nhiều !

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

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

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

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

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

\(\Rightarrow x+2010=0\)

\(\Rightarrow x=-2010\)

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

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

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

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

\(=>x+2010=0\)

\(=>x=-2010\)

\(pt\Leftrightarrow\frac{x}{2009}+\frac{1}{2009}+\frac{x}{2008}+\frac{2}{2008}=\frac{x}{3}+\frac{2007}{3}+\frac{x}{4}+\frac{2006}{4}\Leftrightarrow\frac{x}{2009}+\frac{x}{2008}-\frac{x}{3}-\frac{x}{4}=\frac{2006}{4}+\frac{2007}{3}-\frac{1}{1008}-\frac{1}{2009}\Leftrightarrow x\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{3}-\frac{1}{4}\right)=\frac{2006}{4}+\frac{2007}{3}-\frac{1}{1008}-\frac{1}{2009}\Leftrightarrow x=\frac{\frac{2006}{4}+\frac{2007}{3}-\frac{1}{1008}-\frac{1}{2009}}{\frac{1}{2009}+\frac{1}{2008}-\frac{1}{3}-\frac{1}{4}}=-2010\)