(x+4/2000 + 1)+(x+3/2001 + 1) = (x+2/2002 + 1)+(x+1/2003)+1

(x+2004/2000) + (x+2004/2001) = (x+2004/2002) + (x+2004/2003)

(x+2004).(1/2000+1/2001) = (x+2004).(1/2002+1/2003)

+ Với x+2004=0 suy ra x=-2004. Ta có 0.(1/2000+1/2001)=0.(1/2002+1/2003), đúng

+ Với x+2004 khác 0 thì (x+2004).(1/2000+1/2001) = (x+2004).(1/2002+1/2003)

                                            1/2000+1/2001        =           1/2002+1/2003, vô lí vì 1/2000+1/2001 > 1/2002+1/2003

Vậy x=-2004

đăng hoài thế!!!

67578579875645

ko bt thì đăng

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

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

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

<=> x + 2004 = 0

<=> x = -2004.

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

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

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

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

\(\Rightarrow x=-2004\)

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

\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+20004}{2003}\)

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

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

\(\Leftrightarrow x+2004=0\)

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

Vậy \(x=-2004\)

\(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)

\(\frac{8\left(x+2007\right)}{8\left(x+2000\right)\left(x+2007\right)}-\frac{8\left(x+2000\right)}{8\left(x+2000\right)\left(x+2007\right)}=\frac{7\left(x+2000\right)\left(x+2007\right)}{8\left(x+2000\right)\left(x+2007\right)}\)

\(8x+8.2007-8x+8.2000=7\left(x^2+4007x+2000.2007\right)\)

\(8.7-7\left(x^2+4007x+2000.2007\right)=0\)

\(7\left(8-x^2-4007x-2000.2007\right)=0\)

\(8-x^2-4007x-2000.2007=0\)

\(x^2+4007x+4013992=0\)

\(\left(x^2+2008x\right)+\left(1999x+4013992\right)=0\)

\(\left(x+2008\right)\left(x+1999\right)=0\)

\(\hept{\begin{cases}x=-2008\\x=-1999\end{cases}}\)

\(\frac{1}{\left(x+2000\right)\left(x+2001\right)}+\frac{1}{\left(x+2001\right)\left(x+2002\right)}+\frac{1}{\left(x+2006\right)\left(x+2007\right)}=\frac{7}{8}\)

\(\frac{1}{x+2000}-\frac{1}{x+2001}+\frac{1}{x+2001}-\frac{1}{x+2002}+...+\frac{1}{x+2006}-\frac{1}{x+2007}=\frac{7}{8}\)

\(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)

phần đầu mk thiếu điều kiện,bn tự bổ sung nha

\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

=> \(\frac{1}{x+2000}-\frac{1}{x+2001}+\frac{1}{x+2001}-\frac{1}{x+2002}+....+\frac{1}{x+2006}-\frac{1}{x+2007}=\frac{7}{8}\)

<=> \(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)

<=> \(\frac{7}{\left(x+2000\right)\left(x+2007\right)}=\frac{7}{8}\Leftrightarrow\left(x+2000\right)\left(x+2007\right)=8\)

=> x = -1999 hoặc x = - 2008

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 \(S=\left\{2015\right\}\)

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

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

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

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

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

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

\(\Leftrightarrow2015-x=0\)

\(\Leftrightarrow x=2015\)

Vậy x=2015

\(\text{ĐKXĐ: }x+1\ne0\text{ và }x-2001\ne0\)

\(\Leftrightarrow x\ne-1\text{ và }x\ne2001\)

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

\(=\frac{\left[x.\left(x+1\right)-2001\left(x+1\right)\right].2001}{\left(x+1\right)\left(x-2001\right).2002}=\frac{\left(x-2001\right)\left(x+1\right).2001}{\left(x+1\right)\left(x-2001\right).2002}=\frac{2001}{2002}\)

a)x=2015

ai hok biết, giải ra giùm

a) \(\frac{x-15}{2000}+\frac{x-14}{2001}+\frac{x-13}{2002}=\frac{x-12}{2003}+2\)

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

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

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

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

\(\Rightarrow x-2015=0\)( vì \(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}\ne0\))

\(\Leftrightarrow x=2015\)

Vậy tập hợp nghiệm \(S=\left\{2015\right\}\)