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

\(\Leftrightarrow\dfrac{x}{2000}-1+\dfrac{x+1}{2001}-1+\dfrac{x+2}{2002}-1+\dfrac{x+3}{2003}-1+\dfrac{x+4}{2004}-1=0\)

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

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

Mà \(\dfrac{1}{2000}+\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}>0\)

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

Vậy x = 2000

\(\dfrac{x-4}{2001}\)- 1 +\(\dfrac{x-3}{2002}\)-1 + \(\dfrac{x-2}{2003}\)-1 =\(\dfrac{x-2003}{2}\)-1 + \(\dfrac{x-2002}{3}\)-1 +\(\dfrac{x-2001}{4}\)-1 <=> \(\dfrac{x-2005}{2001}\)+\(\dfrac{x-2005}{2002}\)+\(\dfrac{x-2005}{2003}\)-\(\dfrac{x-2005}{2}\)-\(\dfrac{x-2005}{3}\)-\(\dfrac{x-2005}{4}\)= 0 <=> (x-2005). (\(\dfrac{1}{2001}\)+\(\dfrac{1}{2002}\)+\(\dfrac{1}{2003}\)-\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)) =0 <=> x-2005=0 ( vì \(\dfrac{1}{2001}\) +\(\dfrac{1}{2002}\) +\(\dfrac{1}{2003}\)- \(\dfrac{1}{2}\) -\(\dfrac{1}{3}\)- \(\dfrac{1}{4}\) khác 0) =>x = 2005

x-4/2001+ x-3/2002 + x-2/2003= x-2003/2 + x-2002/3 + x-2001/4

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

(x-2002/3 -1)+(x-2001/4 -1) =0

<=>x-2005/2001+ x-2005/2002+ x-2005/2003- x-2005/2-

x-2005/3- x-2005/4 =0

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

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

<=>x=2005

Vậy pt có nghiệm là x=2005

\(\dfrac{x-4}{2001}+\dfrac{x-3}{2002}+\dfrac{x-2}{2003}=\dfrac{x-2003}{2}+\dfrac{x-2002}{3}+\dfrac{x-2001}{4}\)

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

\(\Leftrightarrow\dfrac{x-2005}{2001}+\dfrac{x-2005}{2002}+\dfrac{x-2005}{2003}-\dfrac{x-2005}{2}-\dfrac{x-2005}{3}-\dfrac{x-2005}{4}=0\)

\(\Leftrightarrow\left(x-2005\right)\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}\ne0\right)=0\)

\(\Leftrightarrow x-2005=0\)

\(\Leftrightarrow x=2005\)

Vậy nghiệm của PT là \(x=2005\)

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

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

\(\Leftrightarrow x=-2005\)

a, \(\dfrac{2-x}{2001}-1=\dfrac{1-x}{2002}-\dfrac{x}{2003}\)

\(\Leftrightarrow\dfrac{2-x}{2001}-1+2=\dfrac{1-x}{2002}-\dfrac{x}{2003}+2\)

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

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

\(\Leftrightarrow\dfrac{2003-x}{2001}-\dfrac{2003-x}{2002}-\dfrac{2003-x}{2003}=0\)

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

Vì \(\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\)

\(\Rightarrow2003-x=0\)

\(\Rightarrow x=2003\)

Vậy : \(s=\left\{2003\right\}\)

b, \(\dfrac{x-5}{100}+\dfrac{x-4}{101}=\dfrac{x-100}{5}+\dfrac{x-101}{4}\)

\(\Leftrightarrow\dfrac{x-5}{100}+\dfrac{x-4}{101}-2=\dfrac{x-100}{5}+\dfrac{x-101}{4}-2\)

\(\Leftrightarrow\left(\dfrac{x-5}{100}-1\right)+\left(\dfrac{x-4}{101}-1\right)=\left(\dfrac{x-100}{5}-1\right)+\left(\dfrac{x-101}{4}-1\right)\)

\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}=\dfrac{x-105}{5}+\dfrac{x-105}{4}\)

\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}-\dfrac{x-105}{5}-\dfrac{x-105}{4}=0\)

\(\Leftrightarrow\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}-\dfrac{1}{5}-\dfrac{1}{4}\right)=0\)

Vì \(\dfrac{1}{100}+\dfrac{1}{101}-\dfrac{1}{5}-\dfrac{1}{4}\ne0\)

\(\Rightarrow x-105=0\)

\(\Rightarrow x=105\)

Vậy : \(s=\left\{105\right\}\)

\(a,\dfrac{2-x}{2001}-1=\dfrac{1-x}{2002}-\dfrac{x}{2003}\)

\(\Leftrightarrow\)haizzz bạn cộng mỗi hạng tử ở mỗi vế cho một. Chuyển vế và giải ra x=2003

b, Tương tự bạn -1 cho mỗi vế. GIải phương trình đc x=105

h.

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

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

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

\(\Leftrightarrow\dfrac{2004-x}{2002}-\dfrac{2004-x}{2003}-\dfrac{2004-x}{2004}=0\)

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

Vì: \(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\ne0\)

Suy ra: 2004 - x = 0

Vậy x = 2004

\(a,\dfrac{x-23}{24}+\dfrac{x-23}{25}=\dfrac{x-23}{26}+\dfrac{x-23}{27}\)

\(\Leftrightarrow\dfrac{x-23}{24}+\dfrac{x-23}{25}-\dfrac{x-23}{26}-\dfrac{x-23}{27}=0\)

\(\Leftrightarrow\left(x-23\right)\left(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\right)=0\)

\(\Leftrightarrow x-23=0\) ( vì \(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\ne0\) )

\(\Leftrightarrow x=23\)

Vậy pt có tập nghiệm S = { 23 }

\(b,\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)

\(\Leftrightarrow\dfrac{x+2+98}{98}+\dfrac{x+3+97}{97}-\dfrac{x+4+96}{96}-\dfrac{x+5+95}{95}=0\)

\(\Leftrightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}-\dfrac{x+100}{96}-\dfrac{x+100}{95}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)

\(\Leftrightarrow x+100=0\)

\(\Leftrightarrow x=-100\)

Vậy pt có tập nghiệm S = { 100 }

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

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

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

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

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

\(\Leftrightarrow x+2005=0\)

\(\Leftrightarrow x=-2005\)

Vậy pt có tập nghiệm S = { 2005 }

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

\(\Leftrightarrow\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)

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

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

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

\(\Leftrightarrow300-x=0\)

\(\Leftrightarrow x=300\)

Vậy pt có tập nghiệm S = { 300 }

\(e,\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53}{47}\)

\(\Leftrightarrow\dfrac{x-45}{55}-1+\dfrac{x-47}{53}-1=\dfrac{x-55}{45}-1+\dfrac{x-53}{47}-1\)

\(\Leftrightarrow\dfrac{x-45-55}{55}+\dfrac{x-47-53}{53}-\dfrac{x-55-45}{45}-\dfrac{x-53-47}{47}=0\)

\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}-\dfrac{x-100}{45}-\dfrac{x-100}{47}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\right)=0\)

\(\Leftrightarrow x-100=0\)

\(\Leftrightarrow x=100\)

Vậy pt có tập nghiệm S = { 100 }

\(f,\dfrac{x+1}{9}+\dfrac{x+2}{8}=\dfrac{x+3}{7}+\dfrac{x+4}{6}\)

\(\Leftrightarrow\dfrac{x+1}{9}+1+\dfrac{x+2}{8}+1=\dfrac{x+3}{7}+1+\dfrac{x+4}{6}+1\)

\(\Leftrightarrow\dfrac{x+10}{9}+\dfrac{x+10}{8}-\dfrac{x+10}{7}-\dfrac{x+10}{6}=0\)

\(\Leftrightarrow\left(x+10\right)\left(\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{7}-\dfrac{1}{6}\right)=0\)

\(\Leftrightarrow x+10=0\)

\(\Leftrightarrow x=-10\)

Vậy pt có tập nghiệm S = { 10 }

\(h,\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)

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

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

\(\Leftrightarrow\dfrac{2-x+2002}{2002}-\dfrac{1-x+2003}{2003}-\dfrac{2004-x}{2004}=0\)

\(\Leftrightarrow\dfrac{2004-x}{2002}-\dfrac{2004-x}{2003}-\dfrac{2004-x}{2004}=0\)

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

\(\Leftrightarrow2004-x=0\)

\(\Leftrightarrow x=2004\)

Vậy pt có tập nghiệm S = { 2004 }

\(g,\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+\dfrac{x+8}{92}\)

\(\Leftrightarrow\dfrac{x+2}{98}+1+\dfrac{x+4}{96}+1=\dfrac{x+6}{94}+1+\dfrac{x+8}{92}+1\)

\(\Leftrightarrow\dfrac{x+100}{98}+\dfrac{x+100}{96}-\dfrac{x+100}{94}-\dfrac{x+100}{92}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\right)=0\)

\(\Leftrightarrow x+100=0\)

\(\Leftrightarrow x=-100\)

Vậy pt có tập nghiệm S = { -100 }

a.

\(\dfrac{x-23}{24}+\dfrac{x-23}{25}=\dfrac{x-23}{26}+\dfrac{x-23}{27}\)

\(\Leftrightarrow\left(x-23\right)\left(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\right)=0\)

Vì: \(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\ne0\)

Suy ra x - 23 = 0

\(\Leftrightarrow x=23\)

a, \(\dfrac{x-45}{55}-1+\dfrac{x-47}{53}-1=\dfrac{x-55}{45}-1+\dfrac{x-53}{47}-1\)

\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}=\dfrac{x-100}{45}+\dfrac{x-100}{47}\)

\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\ne0\right)=0\Leftrightarrow x=100\)

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

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

\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\ne0\right)=0\Leftrightarrow x=-2005\)

a. lấy mỗi phân số e cộng vs 2 là bt làm ra liền

b,  - 1 hoặc + 1 vs mỗi phân số nha

Câu a)

b) x-45/55 + x-47/53 = x-55/45 + x-53/47
<=>x-45/55 -1 + x-47/53 -1= x-55/45 -1 + x-53/47 - 1
<=>x-100/55 + x-100/53 = x-100/45 + x-100/47
<=>(x-100)(1/55+1/53-1/45-1/47)=0
<=>x-100=0
<=>x=100

Vậy x = 100

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

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

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

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

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

\(\Leftrightarrow x+2005=0\)

\(\Leftrightarrow x=-2005\)

Vậy pt có tập nghiệm S = { 2005 }

\(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}=\dfrac{x-5}{2004}+\dfrac{x-4}{2005}+\dfrac{x-3}{2006}\)

\(\Leftrightarrow\left(\dfrac{x-8}{2001}+1\right)+\left(\dfrac{x-7}{2002}+1\right)+\left(\dfrac{x-6}{2003}+1\right)=\left(\dfrac{x-5}{2004}+1\right)+\left(\dfrac{x-4}{2005}+1\right)+\left(\dfrac{x-3}{2006}+1\right)\)

\(\Leftrightarrow\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}-\dfrac{x-2009}{2004}-\dfrac{x-2009}{2005}-\dfrac{x-2009}{2006}=0\)

\(\Leftrightarrow\left(x-2009\right).\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}-\dfrac{1}{2006}\right)=0\)

\(\text{Mà}:\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}-\dfrac{1}{2006}\right)\ne0\)

\(\Rightarrow x-2009=0\Rightarrow x=2009\)

\(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}=\dfrac{x-5}{2004}+\dfrac{x-4}{4}+\dfrac{x-5}{2006}\)

\(\Leftrightarrow\left(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}\right)-3=\left(\dfrac{x-5}{2004}+\dfrac{x-4}{4}+\dfrac{x-5}{2006}\right)-3\)

\(\Leftrightarrow\left(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}\right)-\left(1+1+1\right)=\left(\dfrac{x-5}{2004}+\dfrac{x-4}{2005}+\dfrac{x-5}{2006}\right)-\left(1+1+1\right)\)

\(\Leftrightarrow\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}-1-1-1=\dfrac{x-5}{2004}+\dfrac{x-4}{2005}+\dfrac{x-5}{2006}-1-1-1\)

\(\Leftrightarrow\left(\dfrac{x-8}{2001}-1\right)+\left(\dfrac{x-7}{2002}-1\right)+\left(\dfrac{x-6}{2003}-1\right)=\left(\dfrac{x-5}{2004}-1\right)+\left(\dfrac{x-4}{2005}-1\right)+\left(\dfrac{x-5}{2006}-1\right)\)

\(\)\(\Leftrightarrow\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}=\dfrac{x-2009}{2004}+\dfrac{x-2009}{2006}+\dfrac{x-2009}{2006}\)

\(\Leftrightarrow\left(\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}\right)-\left(\dfrac{x-2009}{2004}+\dfrac{x-2009}{2006}+\dfrac{x-2009}{2006}\right)=0\)

\(\Leftrightarrow\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}-\dfrac{x-2009}{2004}-\dfrac{x-2009}{2006}-\dfrac{x-2009}{2006}=0\)

\(\Leftrightarrow\left(x-2009\right)\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}-\dfrac{1}{2006}\right)=0\)

\(\Leftrightarrow x-2009=0\)

\(\Leftrightarrow x=2009\)

Vậy \(x=2009\)