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15 tháng 4 2017

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

\(\Leftrightarrow\frac{x+2005}{2004}+\frac{x+2005}{2004}-\frac{x+2005}{2003}-\frac{x+2005}{2003}=0\)

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

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

15 tháng 4 2017

=> (x+1)/2004+1+(x+2)/2003+1=(x+3)/2002+1+(x+4)/2001+1
=> (x+2005)/2004+(x+2005)/2003=(x+2005)/2002+(x+2005)/2001
=> (x+2005)(1/2004+1/2003-1/2002-1/2001)=0
=> x+2005=0
=> x=-2005

6 tháng 7 2019

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

\(\Leftrightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)

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

\(\Leftrightarrow x+2005=0\)

\(\Leftrightarrow x=-2005\)

b) Sửa đề :

\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\)

\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)

\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)

\(\Leftrightarrow x=300\)

c) \(\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\)

\(\Leftrightarrow\frac{2-x}{2002}+1=\frac{1-x}{2003}+1-\frac{x}{2004}+1\)

\(\Leftrightarrow\frac{2004-x}{2002}=\frac{2004-x}{2003}-\frac{2004-x}{2004}\)

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

\(\Leftrightarrow x=2004\)

Vậy....

7 tháng 3 2018

Bạn chuyển về 1 vế sau đó trừ 1 vào mỗi phân thức ta được : 

\(\left(x-2005\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}-\frac{1}{2005}\right)=0\)

Vì biểu thức bên phải khác 0 nên : \(x-2005=0\)=> \(x=2005\)

23 tháng 3 2020

\(\frac{x-5}{2000}+\frac{x-4}{2001}+\frac{x-3}{2002}=\frac{x-2}{2003}+\frac{x-1}{2004}+\frac{x}{2005}\)

\(\Leftrightarrow\frac{x-2005}{2000}+\frac{x-2005}{2001}+\frac{x-2005}{2002}=\frac{x-2005}{2003}+\frac{x-2005}{2004}+\frac{x-2005}{2005}\)

\(\Leftrightarrow\left(x-2005\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}-\frac{1}{2005}\right)=0\)

<=> x - 2005 = 0

<=> x = 2005

Vậy ...............

23 tháng 9 2016

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

=> \(\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}-\frac{x-4}{2001}=0\)

=> \(\left(\frac{x-1}{2004}-1\right)+\left(\frac{x-2}{2003}-1\right)-\left(\frac{x-3}{2002}-1\right)-\left(\frac{x-4}{2001}-1\right)=0\)

=> \(\frac{x-2005}{2004}+\frac{x-2005}{2003}-\frac{x-2005}{2002}-\frac{x-2005}{2001}=0\)

=> \(\left(x-2005\right).\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)

Vì \(\frac{1}{2004}< \frac{1}{2002}\)\(\frac{1}{2003}< \frac{1}{2001}\)

=> \(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\ne0\)

=> \(x-2005=0\)

=> \(x=2005\)

Vậy \(x=2005\)

7 tháng 2 2018

Ta có: \(\frac{x-1}{2004}+\frac{x-2}{2003}=\frac{x-3}{2002}+\frac{x-4}{2001}\)

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

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

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

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

Vì \(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\ne0\)

=> x - 2005 = 0

=> x             = 2005

Vậy x = 2005

=> (x - 1)/2004 - 1 + (x - 2)/2003 - 1 = (x - 3)/2002 -1 + (x - 4)/2001 - 1

=> (x - 2005)/2004 + (x - 2005)/2003 = (x - 2005)/2002 + (x - 2005)/2001

=> (x - 2005)/2004 + (x - 2005)/2003 - (x - 2005)/2002 - (x - 2005)/2001 = 0

=> (x - 2005) * ( 1/2004 + 1/2003 - 1/2002 - 1/2001) = 0

Ta thấy  ( 1/2004 + 1/2003 - 1/2002 - 1/2001) khác 0

=> x - 2005 = 0

=> x = 2005

     

1 tháng 2 2018

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

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

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

\(\Leftrightarrow\)\(\frac{x-2005}{2004}+\frac{x-2005}{2003}\)\(=\frac{x-2015}{2001}+\frac{x-2005}{2002}\)

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

\(\Leftrightarrow\)( x - 2005 ) ( \(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2001}-\frac{1}{2002}\))  =  0

Do  \(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2001}-\frac{1}{2002}\)\(\ne\)0

\(\Rightarrow\)x  -   2005   =  0

\(\Leftrightarrow\)x  =  2005

Vậy  x  =  2005

2 tháng 3 2018

2.

pt <=> (x/2000 - 1) + (x+1/2001 - 1) + (x+2/2002 - 1) + (x+3/2003 - 1) + (x+4/2004 - 1 ) = 0

<=> x-2000/2000 + x-2000/2001 + x-2000/2002 + x-2000/2003 + x-2000/2004 = 0

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

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

<=> x=2000

Tk mk nha

2 tháng 3 2018

1.

a, = (2x-1)^2-2.(2x-1)+1-4

    = (2x-1-1)^2-4

    = (2x-2)^2-4

    = (2x-2-2).(2x-2+2)

    = 2x.(2x-4)

b, = [x.(x+3)].[(x+1).(x+2)]

    = (x^2+3x).(x^2+3x+1)-8

    = (x^2+3x+1)^2-1-8

    = (x^2+3x+1)^2-9

    = (x^2+3x+1-3).(x^2+3x+1+3)

    = (x^2+3x-2).(x^2+3x+4)

    = ((x+1).(x+3).(x^2+3x-2)

Tk mk nha

19 tháng 11 2019

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

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

\(\Leftrightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)

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

De thay \(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}< 0\Rightarrow x+2005=0\)

\(\Rightarrow x=-2005\)

19 tháng 11 2019

                              Bài giải

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

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

\(\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)

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

\(\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)

Do : \(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\ne0\) 

\(\Rightarrow\text{ }x+2005=0\)

\(x=0-2005\)

\(x=-2005\)