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 .

minh lam duoc roi . cach viet phan so ban bam vao o mau vang o cuoi trang .cu di con chuot xuong cuoi trang thi thay 1 o vang , vao xem huong dan la biet ngay ma.

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

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

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

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

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

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

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

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

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

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

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

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

\(\Rightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=\left(x+2009\right)\left(\frac{1}{2005}+\frac{2}{2004}+\frac{1}{2003}\right)\)

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

\(\Rightarrow x+2009=0\)( Vì \(\frac{1}{2008}+\frac{1}{207}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\ne0\))

=> x = -2009

tick cho mình rồi mình làm cho

=1+1/2001+1+1/2002+1+1/2003+...+1+1/2008=8+1/2001+1/2002+1/2003+...+1/2008>8

\(\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>8\)

Ta có:

2002/2001=1+1/2001

2003/2002=1+1/2002

2004/2003= 1+ 1/2003

2005/2004= 1+ 1/2004

2006/2005=1+ 1/2005

2007/2006= 1+ 1/2006

2008/2007=1 + 1/2007.

2009/2008=1+ 1/2008.

=> 2002/2001+2003/2002+2004?2003+2005/2004+2006/2005+ 2007/2006+ 2008/2007+ 2009/2008= 1+1+1+1+1+1+1+1+1/2001+1/2002+1/2003+1/2004+1/2005+1/2006+1/2007+1/2008>8.

Nhớ k đúng cho mình nha!! Thanks!!!

a)  \(\Leftrightarrow\frac{x+7}{2003}+1+\frac{x+4}{2006}+1-\frac{x-1}{2011}-1-\frac{x-5}{2015}-1=0\)

     \(\Leftrightarrow\frac{x+2010}{2003}+\frac{x+2010}{2006}-\frac{x+2010}{2011}-\frac{x+2010}{2015}=0\)

     \(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2003}+\frac{1}{2006}-\frac{1}{2011}-\frac{1}{2015}\right)=0\)

     \(\Leftrightarrow x+2010=0\) ( vì 1/2003  +  1/2006  --  1/2011  -- 1/2015   \(\ne\)0)

    \(\Leftrightarrow x=-2010\)

câu b làm tương tự (có gì không hiểu hỏi mk nha) >v<

1) Ta có : \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)

=> x + 1 = 0

=> x = - 1

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

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

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

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

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

=> x + 2010 = 0

=> x = -2010

c) \(\frac{x+1945}{45}+\frac{x+1954}{54}=\frac{x+1975}{75}+\frac{x+1969}{69}\)

\(\Rightarrow\left(\frac{x+1945}{45}-1\right)+\left(\frac{x+1954}{54}-1\right)=\left(\frac{x+1975}{75}-1\right)+\left(\frac{x+1969}{69}-1\right)\)

=> \(\frac{x+1900}{45}+\frac{x+1900}{54}=\frac{x+1900}{75}+\frac{x+1900}{69}\)

=> \(\left(x+1900\right)\left(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\right)=0\)

=> \(x+1900=0\left(\text{Vì }\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\ne0\right)\)

=> x = -1900

d) \(\frac{x+2008}{10}+\frac{x+2010}{9}=\frac{x+2012}{8}+\frac{x+2014}{7}\)

=> \(\left(\frac{x+2008}{10}+2\right)+\left(\frac{x+2010}{9}+2\right)=\left(\frac{x+2012}{8}+2\right)+\left(\frac{x+2014}{7}+2\right)\)

=> \(\frac{x+2028}{10}+\frac{x+2028}{9}=\frac{x+2028}{8}+\frac{x+2028}{7}\)

=> \(\left(x+2028\right)\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\right)=0\)

=> x + 2028 = 0 \(\left(\text{Vì }\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\ne0\right)\)

=> x = -2028

1) Ta có: \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

        \(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

        \(\Leftrightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

  + TH₁: \(x+1=0\)\(\Leftrightarrow\)\(x=-1\)

  + TH₂: \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}=0\)

      Vì \(\hept{\begin{cases}\frac{1}{10}>\frac{1}{13}\\\frac{1}{11}>\frac{1}{14}\\\frac{1}{12}>0\end{cases}}\)\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}>\frac{1}{13}+\frac{1}{14}\)

            \(\Rightarrow\)\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0\)

             mà \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}=0\)

             \(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=-1\)

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

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

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

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

  + TH₁: \(x+2010=0\)\(\Leftrightarrow\)\(x=-2010\)

  + TH₂: \(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}=0\)

      Vì \(\hept{\begin{cases}\frac{1}{2006}>\frac{1}{2008}\\\frac{1}{2007}>\frac{1}{2009}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{2006}+\frac{1}{2007}>\frac{1}{2008}+\frac{1}{2009}\)

              \(\Rightarrow\)\(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}>0\)

               mà \(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}=0\)

               \(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=-2010\)

3) Ta có: \(\frac{x+1945}{45}+\frac{x+1954}{54}=\frac{x+1975}{75}+\frac{x+1969}{69}\)

        \(\Leftrightarrow\left(\frac{x+1945}{45}-1\right)+\left(\frac{x+1954}{54}-1\right)-\left(\frac{x+1975}{75}-1\right)-\left(\frac{x+1969}{69}-1\right)=0\)

        \(\Leftrightarrow\frac{x+1900}{45}+\frac{x+1900}{54}-\frac{x+1900}{75}-\frac{x+1900}{69}=0\)

       \(\Leftrightarrow\left(x+1900\right).\left(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\right)=0\)

  + TH₁: \(x+1900=0\)\(\Leftrightarrow\)\(x=-1900\)

  + TH₂: \(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}=0\)

      Vì \(\hept{\begin{cases}\frac{1}{45}>\frac{1}{75}\\\frac{1}{54}>\frac{1}{69}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{45}+\frac{1}{54}>\frac{1}{75}+\frac{1}{69}\)

              \(\Rightarrow\)\(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}>0\)

               mà \(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}=0\)

               \(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=-1900\)

4) Ta có: \(\frac{x-99}{5}+\frac{x-97}{7}=\frac{x-95}{9}+\frac{x-93}{11}\)

         \(\Leftrightarrow\left(\frac{x-99}{5}-1\right)+\left(\frac{x-97}{7}-1\right)-\left(\frac{x-95}{9}-1\right)-\left(\frac{x-93}{11}-1\right)=0\)

         \(\Leftrightarrow\frac{x-104}{5}+\frac{x-104}{7}-\frac{x-104}{9}-\frac{x-104}{11}=0\)

         \(\Leftrightarrow\left(x-104\right).\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)=0\)

  + TH₁: \(x-104=0\)\(\Leftrightarrow\)\(x=104\)

  + TH₂: \(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}=0\)

      Vì \(\hept{\begin{cases}\frac{1}{5}>\frac{1}{7}\\\frac{1}{9}>\frac{1}{11}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{5}+\frac{1}{7}>\frac{1}{9}+\frac{1}{11}\)

              \(\Rightarrow\)\(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}>0\)

               mà \(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}=0\)

               \(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=104\)

5) Ta có: \(\frac{x+2008}{10}+\frac{x+2010}{9}=\frac{x+2012}{8}+\frac{x+2014}{7}\)

        \(\Leftrightarrow\left(\frac{x+2008}{10}+2\right)+\left(\frac{x+2010}{9}+2\right)-\left(\frac{x+2012}{8}+2\right)-\left(\frac{x+2014}{7}+2\right)=0\)

        \(\Leftrightarrow\frac{x+2028}{10}+\frac{x+2028}{9}-\frac{x+2028}{8}-\frac{x+2028}{7}=0\)

        \(\Leftrightarrow\left(x+2028\right).\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\right)=0\)

    + TH₁: \(x+2028=0\)\(\Leftrightarrow\)\(x=-2028\)

    + TH₂: \(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}=0\)

      Vì \(\hept{\begin{cases}\frac{1}{10}< \frac{1}{8}\\\frac{1}{9}< \frac{1}{7}\end{cases}}\)\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{9}< \frac{1}{8}+\frac{1}{7}\)

              \(\Rightarrow\)\(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}< 0\)

               mà \(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}=0\)

               \(\Rightarrow\)Phương trình trên vô nghiệm

Vậy \(x=-2028\)

Chúc bn hok tốt nha

1 . \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Vì : \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)

\(\Rightarrow x+1=0\)

\(\Rightarrow x=-1\)

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

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

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

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

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

\(\Rightarrow x+2010=0\)

\(\Rightarrow x=-2010\)