`Answer:`

\(\left(\frac{x+1}{2013}\right)+\left(\frac{x+2}{2012}\right)+\left(\frac{x+3}{2011}\right)=\left(\frac{x+4}{2010}\right)+\left(\frac{x+5}{2009}\right)+\left(\frac{x+6}{2008}\right)\)

\(\Leftrightarrow\frac{x+1}{2013}+1+\frac{x+2}{2012}+1+\frac{x+3}{2011}+1=\frac{x+4}{2010}+1+\frac{x+5}{2009}+1+\frac{x+6}{2008}+1\)

\(\Leftrightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}=\frac{x+2014}{2010}+\frac{x+2014}{2009}+\frac{x+2014}{2008}\)

\(\Leftrightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}-\frac{x+2014}{2010}-\frac{x+2014}{2009}-\frac{x+2014}{2008}=0\)

\(\Leftrightarrow\left(x+2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)

\(\Rightarrow x+2014=0\)

\(\Leftrightarrow x=-2014\)

trừ 1 vào mỗi tỉ số,ta đc:

\(\frac{x-1}{2011}-1+\frac{x-2}{2010}-1-\frac{x-3}{2009}-1=\frac{x-4}{2008}-1\)

\(\Rightarrow\frac{x-1-2011}{2011}+\frac{x-2-2010}{2010}-\frac{x-3-2009}{2009}=\frac{x-4-2008}{2008}\)

\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}=\frac{x-2012}{2008}\)

\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)

\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)

\(mà\frac{1}{2011}<\frac{1}{2010}<\frac{1}{2009}<\frac{1}{2008}\Rightarrow\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)

=>x-2012=0

=>x=2012

vậy x=2012

sao cộng mà ko trừ đi

bn -3 đi mà hiệu chỉ có trừ 1 à

\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x.\left(x+1\right)}=\frac{2008}{2010}.\)

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

\(\frac{1}{2}-\frac{1}{x+1}=\frac{502}{1005}\)

\(\frac{1}{x+1}=\frac{1}{2010}\)

=> x + 1 = 2010

=> x = 2009

Ta có : \(\frac{2}{2\times3}+\frac{2}{3\times4}+....+\frac{2}{x\times\left(x+1\right)}=\frac{2008}{2010}\)

\(\Rightarrow2\times\left(\frac{1}{2\times3}+.....+\frac{1}{x\times\left(x+1\right)}\right)=\frac{1004}{1005}\)

\(\Rightarrow2\times\left(\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1004}{1005}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1004}{1005}:2\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{502}{1005}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{502}{1005}=\frac{1}{2010}\)

\(\Rightarrow x+1=2010\)

\(\Rightarrow x=2010-1=2009\)

Tag thầy Lâm không :)???

\(3.\)

\(\frac{x-1}{2011}+\frac{x-2}{2010}+\frac{x-3}{2009}=\frac{x-4}{2008}\)

\(\Rightarrow\)\(\frac{x-1}{2011}-1+\frac{x-2}{2010}-1+\frac{x-3}{2009}-1-\frac{x-4}{2008}+1+2=0\)

\(\Rightarrow\)\(\frac{x-1}{2011}-\frac{2011}{2011}+\frac{x-2}{2010}-\frac{2010}{2010}+\frac{x-3}{2009}-\frac{2009}{2009}-\frac{x-4}{2008}+\frac{2008}{2008}=0\)

\(\Rightarrow\)\(\frac{x-2012}{2011}+\frac{x-2012}{2010}+\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)

\(\Rightarrow\)\(x-2012\left(\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}\right)=0\)

\(\Rightarrow\)\(x=2012\)

a,  \((\frac{3}{7}-\frac{2}{3})\)  .x  =\(\frac{10}{21}\)                                                                   

  \(\frac{-5}{21}\).x=\(\frac{10}{21}\)

x= -2

 Mk chỉ làm 1 phần  các phằn còn lại tương tự

Bài rút gọn biểu thức

a) M=|2x-3|+|x-1| với x > 1,5

b) N=|2-x|-3|x+1| với x < -1

c) P=|3x-5|+|x-2|

d) Q=|x-3|-2.|-5x|

a, \(\frac{3}{7}x-\frac{2}{3}x=\frac{10}{21}\Rightarrow\left(\frac{3}{7}-\frac{2}{3}\right)x=\frac{10}{21}\Rightarrow x=\frac{10}{21}.\left(\frac{-21}{5}\right)\Rightarrow x=-2\)

b, \(\frac{7}{35}:\left(x-\frac{1}{3}\right)=\frac{2}{25}\Rightarrow x-\frac{1}{3}=\frac{1}{5}:\frac{2}{25}\Rightarrow x=\frac{1}{10}-\frac{1}{3}=\frac{13}{30}\)

c, \(|2x-4|+1=5\)

\(\Rightarrow\orbr{\begin{cases}2x-4=4\\2x-4=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=0\end{cases}}\)

d, \(3.|3-2x|-1=\frac{2}{5}\)

\(\Rightarrow3.|3-2x|=\frac{7}{5}\)

\(\Rightarrow\orbr{\begin{cases}3-2x=\frac{7}{15}\\3-2x=-\frac{7}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{19}{15}\\x=\frac{26}{15}\end{cases}}\)

f, \(\left(2x-1\right).\left(x+\frac{2}{3}\right)=0\)

\(\Rightarrow\hept{\begin{cases}2x-1=0\\x+\frac{2}{3}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\x=-\frac{2}{3}\end{cases}}\)

(x-4)/2007 + (x-3)/2008)= (x-2)/2009 + (x-1)/2010 
=[(x-4)/2007 -1]+[(x-3)/2008 -1]=[(x-2)/2009 -1]+[(x-1)/2010 -1] 
=(x-2011)/2007+(x-2011)/2008=(x-2011)/... 
=(x-2011)(1/2007+1/2008-1/2009-1/2010)... 
suy ra x=2010 

Ta có:

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

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

\(\left(x+2012\right)\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)

\(x=-2012\)

;Ko tồn tại nghiệm số thực OK