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5 tháng 8 2017

\(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}=\dfrac{x-5}{2004}+\dfrac{x-4}{2005}+\dfrac{x-3}{2006}\)

\(\Leftrightarrow\left(\dfrac{x-8}{2001}+1\right)+\left(\dfrac{x-7}{2002}+1\right)+\left(\dfrac{x-6}{2003}+1\right)=\left(\dfrac{x-5}{2004}+1\right)+\left(\dfrac{x-4}{2005}+1\right)+\left(\dfrac{x-3}{2006}+1\right)\)

\(\Leftrightarrow\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}-\dfrac{x-2009}{2004}-\dfrac{x-2009}{2005}-\dfrac{x-2009}{2006}=0\)

\(\Leftrightarrow\left(x-2009\right).\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}-\dfrac{1}{2006}\right)=0\)

\(\text{Mà}:\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}-\dfrac{1}{2006}\right)\ne0\)

\(\Rightarrow x-2009=0\Rightarrow x=2009\)

6 tháng 8 2017

\(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}=\dfrac{x-5}{2004}+\dfrac{x-4}{4}+\dfrac{x-5}{2006}\)

\(\Leftrightarrow\left(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}\right)-3=\left(\dfrac{x-5}{2004}+\dfrac{x-4}{4}+\dfrac{x-5}{2006}\right)-3\)

\(\Leftrightarrow\left(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}\right)-\left(1+1+1\right)=\left(\dfrac{x-5}{2004}+\dfrac{x-4}{2005}+\dfrac{x-5}{2006}\right)-\left(1+1+1\right)\)

\(\Leftrightarrow\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}-1-1-1=\dfrac{x-5}{2004}+\dfrac{x-4}{2005}+\dfrac{x-5}{2006}-1-1-1\)

\(\Leftrightarrow\left(\dfrac{x-8}{2001}-1\right)+\left(\dfrac{x-7}{2002}-1\right)+\left(\dfrac{x-6}{2003}-1\right)=\left(\dfrac{x-5}{2004}-1\right)+\left(\dfrac{x-4}{2005}-1\right)+\left(\dfrac{x-5}{2006}-1\right)\)

\(\)\(\Leftrightarrow\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}=\dfrac{x-2009}{2004}+\dfrac{x-2009}{2006}+\dfrac{x-2009}{2006}\)

\(\Leftrightarrow\left(\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}\right)-\left(\dfrac{x-2009}{2004}+\dfrac{x-2009}{2006}+\dfrac{x-2009}{2006}\right)=0\)

\(\Leftrightarrow\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}-\dfrac{x-2009}{2004}-\dfrac{x-2009}{2006}-\dfrac{x-2009}{2006}=0\)

\(\Leftrightarrow\left(x-2009\right)\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}-\dfrac{1}{2006}\right)=0\)

\(\Leftrightarrow x-2009=0\)

\(\Leftrightarrow x=2009\)

Vậy \(x=2009\)

16 tháng 7 2023

\(\dfrac{x+5}{2005}+\dfrac{x+6}{2004}+\dfrac{x+7}{2003}=-3\\ \Rightarrow\dfrac{x+5}{2005}+\dfrac{x+6}{2004}+\dfrac{x+7}{2003}+3=0\\ \Rightarrow\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+6}{2004}+1\right)+\left(\dfrac{x+7}{2003}+1\right)=0\\ \Rightarrow\dfrac{x+2010}{2005}+\dfrac{x+2010}{2004}+\dfrac{x+2010}{2003}=0\\ \Rightarrow\left(x+2010\right)\left(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\right)=0\\ \Rightarrow x+2010=0\left(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\ne0\right)\\ \Rightarrow x=-2010\)

5 tháng 10 2021

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)

\(\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)

\(\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)\)

\(x+2014=0\)

\(x=-2014\)

5 tháng 10 2021

\(\Rightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\\ \Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\\ \Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\\ \Rightarrow x=-2004\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\right)\)

3 tháng 8 2017

\(\dfrac{x+4}{2001}+\dfrac{x+3}{2002}=\dfrac{x+2}{2003}+\dfrac{x+1}{2004}\)

\(\Leftrightarrow\left(\dfrac{x+4}{2001}+1\right)+\left(\dfrac{x+3}{2002}+1\right)=\left(\dfrac{x+2}{2003}+1\right)+\left(\dfrac{x+1}{2004}+1\right)\)

\(\Leftrightarrow\dfrac{x+2005}{2001}+\dfrac{x+2005}{2002}-\dfrac{x+2005}{2003}-\dfrac{x+2005}{2004}=0\)

\(\Leftrightarrow\left(x+2005\right)\cdot\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)=0\)

\(\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)\ne0\)

\(\Rightarrow x+2005=0\Rightarrow x=-2005\)

25 tháng 9 2021

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}=0\)

<=> \(\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)=0\)

<=> \(\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)

<=> \(\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

<=> x + 2004 = 0

<=> x = -2004

(Bn nhớ thêm kết quả là 0 vào sau nữa nha)

25 tháng 9 2021

camon nhok=))

18 tháng 9 2021

Bài 1:

a) \(\left|3x-5\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)

c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Leftrightarrow x=-2004\)( do \(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\))

Bài 2:

a) \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)

\(=\dfrac{1}{4}+\dfrac{3}{4}=1\)

b) \(=-\left(\dfrac{1}{99.100}+\dfrac{1}{98.99}+\dfrac{1}{97.98}+...+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)\)

\(=-\left(\dfrac{1}{99}-\dfrac{1}{100}+\dfrac{1}{98}-\dfrac{1}{99}+...+1-\dfrac{1}{2}\right)\)

\(=-\left(1-\dfrac{1}{100}\right)=-\dfrac{99}{100}\)

 

18 tháng 9 2021

Bài 1:

a) \(\left|3x-5\right|=4\)  (1)

\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=9\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)

b) \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)

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

\(\Leftrightarrow x+1=0\)    \(\left(do\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\right)\)

\(\Leftrightarrow x=-1\)

c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Leftrightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)

\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Leftrightarrow x+2004=0\)           \(\left(do\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\right)\)

\(\Leftrightarrow x=-2004\)

3 tháng 9 2018

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Leftrightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}\)

\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)

\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Leftrightarrow x+2004=0\)

\(\Leftrightarrow x=-2004\)

3 tháng 9 2018

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)

\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}\right)=\left(x+2004\right)\left(\dfrac{1}{2002}+\dfrac{1}{2003}\right)\)

\(\dfrac{1}{2000}+\dfrac{1}{2001}\ne\dfrac{1}{2002}+\dfrac{1}{2003}\) nên \(x+2004=0\Rightarrow x=-2004\)

Vậy, x = -2004

12 tháng 6 2017

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Rightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Rightarrow x+2004=0\)

\(\Rightarrow x=-2004\)

Vậy x = -2004

12 tháng 6 2017

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)

\(\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)

\(\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

=> x+2004=0 => x=-2004

2 tháng 8 2017

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)

\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Leftrightarrow x+2004=0\Leftrightarrow x=-2004\)

17 tháng 3 2017

\(pt\Leftrightarrow\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+6}{2004}+1\right)+\left(\dfrac{x+7}{2003}+1\right)=0\)

\(\Leftrightarrow\dfrac{x+2010}{2005}+\dfrac{x+2010}{2004}+\dfrac{x+2010}{2003}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\right)=0\)

\(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\ne0\Rightarrow x+2010=0\Rightarrow x=-2010\)

Vậy \(x=-2010\)

17 tháng 3 2017

x = -2010 thì phải