\(\frac{x+4}{2007}+\frac{x+8}{2003}=\frac{x+1}{2010}+\...">
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8 tháng 9 2016

\(\frac{x+4}{2007}+\frac{x+8}{2003}=\frac{x+1}{2010}=\frac{x+3}{2008}\)

\(\Leftrightarrow\frac{x+4}{2007}=\frac{x+1}{2010}\)

\(\Leftrightarrow\left(x+4\right)2010=\left(x+1\right)2007\)

\(\Leftrightarrow2010x+8040=2007x+2007\)

\(\Leftrightarrow2010x-2007x=2007-8040\)

\(\Leftrightarrow3x=-6033\)

\(\Leftrightarrow x=-2011\)

8 tháng 9 2016

\(\frac{x+4}{2007}+\frac{x+8}{2003}=\frac{x+1}{2010}+\frac{x+3}{2008}\)

=>\(\left(\frac{x\text{+4}}{2007}+1\right)+\left(\frac{x+8}{2003}+1\right)=\left(\frac{x+1}{2010}+1\right)+\left(\frac{x+3}{2008}+1\right)\)

=>\(\frac{x+2011}{2007}+\frac{x+2011}{2003}=\frac{x+2011}{2010}+\frac{x+2011}{2008}\)

=>\(\frac{x+2011}{2007}+\frac{x+2011}{2003}-\frac{x+2011}{2010}-\frac{x+2011}{2008}=0\)

=>\(x+2011\left(\frac{1}{2007}+\frac{1}{2003}-\frac{1}{2010}-\frac{1}{2008}\right)=0\)

Mà \(\frac{1}{2007}+\frac{1}{2003}-\frac{1}{2010}-\frac{1}{2008}\ne0\)

=> x+2011=0

=>x=-2011

Vậy x = -2011

10 tháng 5 2019

bên sau là 2 lần -4 à đúng ko đấy ???

10 tháng 5 2019

\(\frac{x+1}{2010}+\frac{x+3}{2008}+\frac{x+4}{2007}+\frac{x+9}{2002}=-4\)

\(\Leftrightarrow\frac{x+1}{2010}+1+\frac{x+3}{2008}+1+\frac{x+4}{2007}+1+\frac{x+9}{2002}+1=-4+4\)

\(\Leftrightarrow\frac{x+2011}{2010}+\frac{x+2011}{2008}+\frac{x+2011}{2007}+\frac{x+2011}{2002}=0\)

\(\Leftrightarrow\left(x+2011\right)\left(\frac{1}{2010}+\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2002}\right)=0\)

\(\Leftrightarrow x+2011=0\)

\(\Leftrightarrow x=-2011\)

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

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

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

\(\Rightarrow x+2009=0\)

\(\Rightarrow x=-2009\)

30 tháng 3 2016

Câu  \(1.\)  Giải phương trình

\(a.\)  \(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)   \(\left(1\right)\)

Đặt  \(y=x^2+x\)  \(\left(2\right)\)  thì khi đó, phương trình  \(\left(1\right)\)  sẽ có dạng:

\(y^2+4y=12\)

\(\Leftrightarrow\)   \(y^2+4y-12=0\)

\(\Leftrightarrow\)   \(y^2+4y+4-16=0\)

\(\Leftrightarrow\)   \(\left(y+2\right)^2-4^2=0\)

\(\Leftrightarrow\)   \(\left(y-2\right)\left(y+6\right)=0\)

\(\Leftrightarrow\)   \(^{y-2=0}_{y+6=0}\)  \(\Leftrightarrow\)  \(^{y=2}_{y=-6}\)      

Đến bước này, ta cần xét hai trường hợp sau:

\(\text{*)}\)  \(TH_1:\)  Với  \(y=2\)  thì phương trình  \(\left(2\right)\)  trở thành:

\(x^2+x=2\)

\(\Leftrightarrow\)  \(x^2+x-2=0\)

\(\Leftrightarrow\)  \(\left(x^2-1\right)+x-1=0\)

\(\Leftrightarrow\)  \(\left(x-1\right)\left(x+1\right)+\left(x-1\right)=0\)

\(\Leftrightarrow\)  \(\left(x-1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\)   \(^{x-1=0}_{x+2=0}\)  \(\Leftrightarrow\)  \(^{x=1}_{x=-2}\)  (dùng dấu ngoặc nhọn nhé bạn!)  

\(\text{*)}\)  \(TH_2:\)  Với  \(y=-6\)  thì phương trình  \(\left(2\right)\)  trở thành:

\(x^2+x=-6\)

\(\Leftrightarrow\)  \(x^2+x+6=0\)

\(\Leftrightarrow\)  \(x^2+2.\frac{1}{2}.x+\frac{1}{4}+\frac{23}{4}=0\)

\(\Leftrightarrow\)  \(\left(x+\frac{1}{2}\right)^2+\frac{23}{4}=0\)  \(\left(3\right)\)

Vì  \(\left(x+\frac{1}{2}\right)^2\ge0\)  với mọi  \(x\)  \(\Rightarrow\)  \(\left(x+\frac{1}{2}\right)^2+\frac{23}{4}\ge\frac{23}{4}>0\)

Do đó, phương trình  \(\left(3\right)\)  vô nghiệm!

Vậy,  tập nghiệm của phương trình  \(\left(1\right)\)  là  \(S=\left\{-1;2\right\}\)

30 tháng 3 2016

Câu  \(1.\)  Giải phương trình!

\(b.\) 

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

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

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

\(\Leftrightarrow\)    \(\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)   \(\left(4\right)\)

Do  \(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\ne0\)  nên từ  \(\left(4\right)\)  suy ra  

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

Vậy,  \(S=\left\{-2009\right\}\)

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

\(\text{Giải}\)

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

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

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