x=-2007

\(a.\left(\frac{x+1}{2000}+1\right)+\left(\frac{x+2}{1999}+1\right)+\left(\frac{x+3}{1998}+1\right)+\left(\frac{x+4}{1997}+1\right)=0\)

\(=\frac{x+2001}{2000}+\frac{x+2001}{1999}+\frac{x+2001}{1998}+\frac{x+2001}{1997}=0\)

\(=\left(x+2001\right).\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\right)=0\)

\(=>x+2001=0\)

\(x=-2001\)

\(b.\left(\frac{x+1}{1999}-1\right)+\left(\frac{x+2}{2000}-1\right)+\left(\frac{x+3}{2001}-1\right)=\left(\frac{x+4}{2002}-1\right)+\left(\frac{x+5}{2003}-1\right)\)\(+\left(\frac{x+6}{2004}-1\right)\)

\(\frac{x+1998}{1999}+\frac{x+1998}{2000}+\frac{x+1998}{2001}=\frac{x+1998}{2002}+\frac{x+1998}{2003}+\frac{x+1998}{2004}\)

\(\frac{x+1998}{1999}+\frac{x+1998}{2000}+\frac{x+1998}{2001}-\frac{x+1998}{2002}-\frac{x+1998}{2003}-\frac{x+1998}{2004}=0\)

\(\left(x+1998\right).\left(\frac{1}{1999}+\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)=0\)

\(=>x+1998=0\)

\(x=-1998\)

dễ quá!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

\(\frac{x+1}{2000}+\frac{x+2}{1999}+\frac{x+3}{1998}+\frac{x+4}{1997}=-4\)

\(\Leftrightarrow\left(\frac{x+1}{2000}+1\right)+\left(\frac{x+2}{1999}+1\right)+\left(\frac{x+3}{1998}+1\right)+\) \(\left(\frac{x+4}{1997}+1\right)=0\)

\(\Leftrightarrow\frac{x+2001}{2000}+\frac{x+2001}{1999}+\frac{x+2001}{1998}+\frac{x+2001}{1997}=0\)

\(\Leftrightarrow\left(x+2001\right)\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\right)=0\)

Mà : \(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\ne0\)

\(\Rightarrow x+2001=0\)

\(\Leftrightarrow x=-2001\)

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....

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

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

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

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

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

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 \(x=2005\)

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

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

bạn có thể giải giùm  mk bài toán hình nx k ạ

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

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

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

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

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

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

\(\Leftrightarrow\)( 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

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

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

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

=> x + 2015 = 0

=> x = -2015

Vậy x = -2015

TL :

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

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

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

\(\Rightarrow x+2005=0\)

\(\Rightarrow x=-2005\)

Chuyển vế rồi tách x+2005 ra nhé