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4 tháng 4 2020

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

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

<=> \(\frac{x-2004}{2000}+\frac{x-2004}{2001}+\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{3}+\frac{x-2004}{4}\)

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

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

<=> x = 2004

Vậy S = {2004}

4 tháng 4 2020

đề bài \(=\frac{x-2002}{2}+\frac{x-2001}{3}+\frac{x-2000}{4}\)

 \(\Leftrightarrow\frac{x}{2000}-\frac{4}{2000}+\frac{x}{2001}-\frac{3}{2001}+\frac{x}{2002}-\frac{2}{2002}=\frac{x}{2}-\frac{2002}{2}+\frac{x}{3}-\frac{2001\\}{3}+\frac{x}{4}-\frac{2000}{4}\)

\(\Leftrightarrow\frac{x}{2000}-\frac{1}{500}+\frac{x}{2001}-\frac{1}{667}+\frac{x}{2002}-\frac{1}{1001}-\frac{x}{2}-\frac{x}{3}-\frac{x}{4}+1001+667+500=0\)

\(\Leftrightarrow\left(\frac{x}{2000}+\frac{x}{2001}+\frac{x}{2002}-\frac{x}{2}-\frac{x}{3}-\frac{x}{4}\right)+\left(1001+667+500-\frac{1}{500}-\frac{1}{667}-\frac{1}{1001}\right)=0\)

=> x=1

14 tháng 2 2019

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

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

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

Với \(x-2004\ne0\)

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

Với \(x-2004=0\)

\(\Rightarrow x=2004\)

8 tháng 2 2017

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

\(\Leftrightarrow\left(\frac{x}{2000}-1\right)+\left(\frac{x+1}{2001}-1\right)+\left(\frac{x+2}{2002}-1\right)=0\)

\(\Leftrightarrow\frac{x-2000}{2000}+\frac{x-2000}{2001}+\frac{x-2000}{2002}=0\)

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

\(\Leftrightarrow x-2000=0\).Do \(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\ne0\)

\(\Leftrightarrow x=2000\)

8 tháng 2 2017

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

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

\(\frac{x-2000}{2000}+\frac{x-2000}{2001}+\frac{x-2000}{2002}=0\)

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

Mà: \(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\ne0\)

\(\Rightarrow\)x-2000=0 

Vậy : x=2000

17 tháng 2 2020

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

\(\Leftrightarrow\left(\frac{x}{2000}-1\right)+\left(\frac{x+1}{2001}-1\right)+\left(\frac{x+2}{2002}-1\right)+\left(\frac{x+3}{2003}-1\right)=4-4=0\)

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

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

\(\Leftrightarrow x-2000=0\)  ( do \(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\ne0\) )

\(\Leftrightarrow x=2000\)

Vậy x = 2000

Đây là cách của lớp 7 nha

@@ Học tốt

17 tháng 2 2020

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

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

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

Do \(\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)\)khác 0

=> \(x-2000=0\)<=> \(x=2000\)

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

8 tháng 2 2017

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

\(\Rightarrow\left(\frac{x}{2000}-1\right)+\left(\frac{x+1}{2001}-1\right)+\left(\frac{x+2}{2002}-1\right)=0\)

\(\Rightarrow\frac{x-2000}{2000}+\frac{x-2000}{2001}+\frac{x-2000}{2002}=0\)

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

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

\(\Rightarrow x-2000=0\)

\(\Rightarrow x=2000\)

Vậy x = 2000

8 tháng 2 2017

Có : \(\frac{x}{2000}+\frac{x+1}{2001}+\frac{x+2}{2002}=3\)

\(\Leftrightarrow\)\(\left(\frac{x}{2000}-1\right)+\left(\frac{x+1}{2001}-1\right)+\left(\frac{x+2}{2002}-1\right)=0\)

\(\Leftrightarrow\) \(\frac{x-2000}{2000}+\frac{x-2000}{2001}+\frac{x-2000}{2002}=0\)

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

\(\Leftrightarrow\) \(x-2000=0\)

( vì \(\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\right)\ne0\) )

\(\Leftrightarrow\) \(x=2000\)

Vậy x = 2000 thì phương trình \(\frac{x}{2000}+\frac{x+1}{2001}+\frac{x+2}{2002}=3\)

28 tháng 8 2016

Ta có :

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

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

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

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

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

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

\(\Rightarrow x+2004=0\)

\(\Rightarrow x=-2004\)

Vậy ...

30 tháng 8 2016

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


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

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

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

ai bít thì giúp mình với nhé

\(a,\frac{15-x}{2000}+\frac{14-x}{2001}=\frac{13-x}{2002}+\frac{12-x}{2003}\)

\(\Leftrightarrow\frac{15-x}{2000}+1+\frac{14-x}{2001}+1=\frac{13-x}{2002}+1+\frac{12-x}{2003}+1\)

\(\Leftrightarrow\frac{15-x+2000}{2000}+\frac{14-x+2001}{2001}=\frac{13-x+2002}{2002}+\frac{12-x+2003}{2003}\)

\(\Leftrightarrow\frac{2015-x}{2000}+\frac{2015-x}{2001}=\frac{2015}{2002}+\frac{2015-x}{2003}\)

\(\Leftrightarrow\left(2015-x\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

mà \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}>0\)

\(\Leftrightarrow2015-x=0\)

\(\Leftrightarrow x=2015\)

KL : PT có nghiệm \(S=\left\{2015\right\}\)

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....