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


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

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


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


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


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


\(\dfrac{1}{2000}\) + \(\dfrac{1}{2001}\) - \(\dfrac{1}{2002}\) - \(\dfrac{1}{2003}\) khác 0


nên x+2004=0

=>x=0-2004
=> x = -2004
vậy S = -2004.

Tick nhabanhqua

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=))

15 tháng 8 2018

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

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

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

<=>  \(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)

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

<=>  \(x+2004=0\)  (do  1/2000 + 1/2001 - 1/2002 - 1/2003 khác 0)

<=>   \(x=-2004\)

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

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

18 tháng 9 2016

Ta có: \(\left(\frac{x+4}{2000}\right)+\left(\frac{x+3}{2001}\right)=\left(\frac{x+2}{2002}\right)+\left(\frac{x+1}{2003}\right)\)

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

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

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

Vì \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}+\frac{1}{2003}\ne0\)

=> x + 2004 =0

=> x             = -2004

22 tháng 7 2019

Xin mạn phép sửa đề:Tìm x biết

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

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

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

Hiển nhiên cái ngoặc to > 0 do đó x + 2004 =0 hay x = -2004

7 tháng 8 2019

ko sửa có lm đc ko

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