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

<=>\(\left(\dfrac{201-x}{99}+1\right)+\left(\dfrac{203-x}{97}+1\right)+\left(\dfrac{205-x}{95}+1\right)=0\)

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

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

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

<=> 300 - x = 0

<=> x = 300

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

<=> \(\dfrac{2-x}{2002}+1=\left(\dfrac{1-x}{2003}+1\right)+\left(\dfrac{x}{2004}+1\right)\){Cộng cả hai vế của phương trình với 2}

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

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

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

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

<=> 2004 - x = 0

<=> x = 2004.

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

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

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

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

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

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

\(\Rightarrow x=2004\) do \(\left(\dfrac{-1}{2002}+\dfrac{1}{2004}+\dfrac{1}{2003}\ne0\right)\)

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

\(\Leftrightarrow x=23\)

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 }

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

\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}=\dfrac{x+100}{96}+\dfrac{x+100}{95}\)

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

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

Vì \(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\ne0\) nên \(x+100=0\Leftrightarrow x=-100\)

b) \(\dfrac{x+1}{1998}+\dfrac{x+2}{1997}=\dfrac{x+3}{1996}+\dfrac{x+4}{1995}\)

\(\Rightarrow\dfrac{x+1}{1998}+1+\dfrac{x+2}{1997}+1=\dfrac{x+3}{1996}+1+\dfrac{x+4}{1995}+1\)

\(\Rightarrow\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}=\dfrac{x+1999}{1996}+\dfrac{x+1999}{1995}\)

\(\Rightarrow\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}-\dfrac{x+1999}{1996}-\dfrac{x+1999}{1995}=0\)

\(\Rightarrow\left(x+1999\right)\left(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\right)=0\)

Vì \(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\ne0\) nên \(x+1999=0\Leftrightarrow x=-1999\)

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

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

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

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

Vì \(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\ne0\) nên \(300-x=0\Leftrightarrow x=300\)

Sửa đề: \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)

Ta có: \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)

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

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

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

mà \(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}>0\)

nên 300-x=0

hay x=300

Vậy:S={300}

`(201-x)/99+(203-x)/97+(205-x)/95+3=0`

`<=>(201-x)/99+1+(203-x)/97+1+(205-x)/95+1=0`

`<=>(300-x)/99+(300-x)/97+(300-x)/95=0`

`<=>(300-x)(1/99+1/97+1/95)=0`

`<=>300-x=0`

`<=>x=300`

Vậy `x=300`

Sai đề.

Vì 205-x/95+3=0

=>205-x/95=0-3=-3

=>205-x=-3*95=-285

=>x=205-(-285)=490

Vậy x=490

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

=>300-x=0

=>x=300

<=> 300-x/99+300-x/97+300-x/95=0

<=>(300-x).(1/99+1/97+1/95)=0

<=>300-x=0

<=>x=300

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

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

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

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

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

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

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

=> \(2003-x=0\)

=> \(x=2003\)

Vậy : S = \(\left\{2003\right\}\)

b, \(\dfrac{2x-3}{97}-\dfrac{2x-4}{96}+\dfrac{2x-5}{95}=\dfrac{2x-6}{94}\)

<=> \(\dfrac{2x-3}{97}-\dfrac{2x-4}{96}=\dfrac{2x-6}{94}-\dfrac{2x-5}{95}\)

<=> \(\dfrac{2x-3}{97}-\dfrac{2x-4}{96}-2=\dfrac{2x-6}{94}-\dfrac{2x-5}{95}-2\)

<=> \(\left(\dfrac{2x-3}{97}-1\right)-\left(\dfrac{2x-4}{96}-1\right)=\left(\dfrac{2x-6}{94}-1\right)-\left(\dfrac{2x-5}{95}-1\right)\)

<=>\(\dfrac{2x-100}{97}-\dfrac{2x-100}{96}=\dfrac{2x-100}{94}-\dfrac{2x-100}{95}\)

<=> \(\dfrac{2x-100}{97}-\dfrac{2x-100}{96}-\dfrac{2x-100}{94}+\dfrac{2x-100}{95}=0\)

<=> \(\left(2x-100\right)\left(\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{94}+\dfrac{1}{95}\right)=0\)

Vì \(\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{94}+\dfrac{1}{95}\ne0\)

=>\(2x-100=0\)

=> \(2x=100\)

=>\(x=50\)

Vậy: S=\(\left\{50\right\}\)