\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2001}{2002}.\frac{2002}{2003}=\frac{1.2.3.....2001.2002}{2.3.4.....2002.2003}=\frac{1}{2003}\)

sai đề rồi ạn ơi ai cùng chung ý nghĩ thì tick

\(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)

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

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

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

\(\Leftrightarrow\) \(x=2003\) 

↔ \(\frac{2-x}{2001}+1\)\(=\left(\frac{1-x}{2002}+1\right)+\left(\frac{x}{2003}+1\right)\)

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

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

↔ x = 2003

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

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

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

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

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

\(\Leftrightarrow x-2000=0\)  ( 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

\(\frac{x}{2000}\)- 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

=> \(x-2000=0\)<=> \(x=2000\)

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

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

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

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

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

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

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

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

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

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

Dễ thấy: \(\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}>0\)

=> y - 2005 = 0

=> y = 2005

Trên thì y - 3 dưới chuyển thành y + 3 v~~