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18 tháng 12 2018

\(\left(\frac{x-7}{2005}-1\right)+\left(\frac{x-6}{2006}-1\right)=\left(\frac{x-5}{2007}-1\right)+\left(\frac{x-4}{2008}-1\right)\)

\(\Leftrightarrow\frac{x-2012}{2005}+\frac{x-2012}{2006}=\frac{x-2012}{2007}+\frac{x-2012}{2008}\)

\(\Leftrightarrow\frac{x-2012}{2005}+\frac{x-2012}{2006}-\frac{x-2012}{2007}-\frac{x-2012}{2008}=0\)

\(\left(x-2012\right).\left(\frac{1}{2005}+\frac{1}{2006}-\frac{1}{2007}-\frac{1}{2008}\right)=0\)

\(\text{vì }\left(\frac{1}{2005}+\frac{1}{2006}-\frac{1}{2007}-\frac{1}{2008}\right)\ne0\Rightarrow x-2012=0\Rightarrow x-2012\)

28 tháng 12 2016

\(\frac{x-7}{2005}+\frac{x-6}{2006}=\frac{x-5}{2007}+\frac{x-4}{2008}\)

\(\Rightarrow\frac{x-7}{2005}-1+\frac{x-6}{2006}-1=\frac{x-5}{2007}-1+\frac{x-4}{2008}-1\)

\(\Rightarrow\frac{x-2012}{2005}+\frac{x-2012}{2006}=\frac{x-2012}{2007}+\frac{x-2012}{2008}\)

\(\Rightarrow\frac{x-2012}{2005}+\frac{x-2012}{2006}-\frac{x-2012}{2007}-\frac{x-2012}{2008}=0\)

\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2005}+\frac{1}{2006}-\frac{1}{2007}-\frac{1}{2008}\right)=0\)

\(\Rightarrow x-2012=0\). Do \(\frac{1}{2005}+\frac{1}{2006}-\frac{1}{2007}-\frac{1}{2008}\ne0\)

\(\Rightarrow x=2012\)

28 tháng 12 2016

\(\frac{x-7}{2005}+\frac{x-6}{2006}=\frac{x-5}{2007}+\frac{x-4}{2008}\)

\(\Rightarrow\left(\frac{x-7}{2005}-1\right)+\left(\frac{x-6}{2006}-1\right)=\left(\frac{x-5}{2007}-1\right)+\left(\frac{x-4}{2008}-1\right)\)

\(\Rightarrow\frac{x-2012}{2005}+\frac{x-2012}{2006}=\frac{x-2012}{2007}+\frac{x-2012}{2008}\)

\(\Rightarrow\frac{x-2012}{2005}+\frac{x-2012}{2006}-\frac{x-2012}{2007}-\frac{x-2012}{2008}=0\)

\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2005}+\frac{1}{2006}-\frac{1}{2007}-\frac{1}{2008}\right)=0\)

\(\left(\frac{1}{2005}+\frac{1}{2006}-\frac{1}{2007}-\frac{1}{2008}\right)\ne0\)

\(\Rightarrow x-2012=0\)

\(\Rightarrow x=2012\)

Vậy \(x=2012\)

13 tháng 11 2015

tick cho mình rồi mình làm cho

16 tháng 11 2015

dễ mà bn,cộng 1 vào mỗi biểu thức và trừ vế 2 là xong

13 tháng 11 2020

\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)

\(\Leftrightarrow\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}+3=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}+3\)

\(\Leftrightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)\)

      \(+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)

\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)

\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)

\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)(1)

Vì \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\ne0\)(2)

Từ (1) và (2) \(\Rightarrow x+2009=0\)\(\Rightarrow x=-2009\)

Vậy \(x=-2009\)

29 tháng 3 2020
https://i.imgur.com/xG3Mq3b.jpg
23 tháng 2 2020

Ta có : \(\frac{x^2-2008}{2007}+\frac{x^2-2007}{2006}+\frac{x^2-2006}{2005}=\frac{x^2-2005}{2004}+\frac{x^2-2004}{2003}+\frac{x^2-2003}{2002}\)

=> \(\frac{x^2-2008}{2007}+1+\frac{x^2-2007}{2006}+1+\frac{x^2-2006}{2005}+1=\frac{x^2-2005}{2004}+1+\frac{x^2-2004}{2003}+1+\frac{x^2-2003}{2002}+1\)

=> \(\frac{x^2-2008}{2007}+\frac{2007}{2007}+\frac{x^2-2007}{2006}+\frac{2006}{2006}+\frac{x^2-2006}{2005}+\frac{2005}{2005}=\frac{x^2-2005}{2004}+\frac{2004}{2004}+\frac{x^2-2004}{2003}+\frac{2003}{2003}+\frac{x^2-2003}{2002}+\frac{2002}{2002}\)

=> \(\frac{x^2-1}{2007}+\frac{x^2-1}{2006}+\frac{x^2-1}{2005}=\frac{x^2-1}{2004}+\frac{x^2-1}{2003}+\frac{x^2-1}{2002}\)

=> \(\frac{x^2-1}{2007}+\frac{x^2-1}{2006}+\frac{x^2-1}{2005}-\frac{x^2-1}{2004}-\frac{x^2-1}{2003}-\frac{x^2-1}{2002}=0\)

=> \(\left(x^2-1\right)\left(\frac{1}{2007}+\frac{1}{2006}+\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\right)=0\)

=> \(x^2-1=0\)

=> \(x^2=1\)

=> \(x=\pm1\)

Vậy phương trình có 2 nghiệm là x = 1, x = -1 .

24 tháng 2 2020

Thanks bn

31 tháng 5 2017

câu 2 :

 \(\Leftrightarrow\)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}-\frac{x+4}{2005}-\frac{x+5}{2004}-\frac{x+6}{2003}\)=0

\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x-2009}{2003}\)=0

\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\)

\(\Rightarrow x+2009=0\)

\(\Rightarrow x=-2009\)

12 tháng 7 2017

c) Ta có : \(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)

\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\)\(\left(\frac{x+6}{2003}+1\right)\)

\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)

\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)

\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)

Mà : \(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\ne0\)

Nên x + 2009 = 0 => x = -2009

30 tháng 1 2019

\(b,\)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)

\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)

\(\Rightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)

\(\Rightarrow\left(x+9\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=\left(x+9\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)\)

\(\Rightarrow\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}=\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\left(KTM\right)\)

30 tháng 1 2019

\(\text{Giải}\)

\(b,\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)

\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)

\(\Leftrightarrow x+2009=0\Leftrightarrow x=-2009\)

18 tháng 6 2016

theo đề baiif nên

x+1/2008+x+2/2007+x+3/2006-(x+4/2005)-(x+5/2004)-(x+6/2003)=0

suy ra [(x+1/2008)+1]+[(x+2/2007)+1]+[x+3/2006)+1]-[(x+4/2005)+1]-[(x+5/2004)+1]-[(x+6/2003)+1]=0

suy ra (x+2009/2008)+(x+2009/2007)+(x+2009/2006)-(x+2009/2005)-(x+2009/2004)-(x+2009/2003)=0

nên (x+2009)(1/2008+1/2007+1/2006-1/2005-1/2004-1/2003)=0

         V1                           V2

Dễ thấy V2>0 NÊN x+2009=0   suy ra x=-2009

26 tháng 11 2017

bit roi hoi nguoi khac lam ji vo duyen