Đề bài tương đương:

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

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

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

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

\(\Leftrightarrow300-x=0\left(\frac{1}{99}+\frac{1}{97}-\frac{1}{95}\ne0\right)\)

\(\Leftrightarrow x=300\)

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}

?!!!

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

\(\Leftrightarrow\frac{201-x}{99}+1+\frac{203-x}{97}+1-\frac{205-x}{95}-1=4\)

\(\Leftrightarrow\frac{200-x}{99}+\frac{200-x}{97}-\frac{200-x}{95}=4\)

\(\Leftrightarrow\left(200-x\right)\left(\frac{1}{99}+\frac{1}{97}-\frac{1}{95}\right)=4\)

Bạn tự làm tiếp.

X = -104,695575 

   Đáp số ra lẻ quá bạn nhỉ 

201-x/99+203-x/97=205-x/95+3

201-x/99+1 +203-x/97+1-205-x/95+1=4

200-x/99+200-x/97+200-x/95=4

(200-x)(1/99+1/97+1/95)=4

a)

\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\\ \Leftrightarrow\frac{201-x}{99}+\frac{99}{99}+\frac{203-x}{97}+\frac{97}{97}+\frac{205-x}{95}+\frac{95}{95}+4=4\\ \Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\) (*)

Do \(\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)\ne0\)

nên (*) \(\Leftrightarrow300-x=0\\ \Leftrightarrow x=300\)

b)

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

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

Do \(\left(\frac{1}{2002}-\frac{1}{2003}+\frac{1}{2004}\right)\ne0\)

nên (*) \(\Leftrightarrow2004-x=0\)

\(\Leftrightarrow x=2004\)

c) \(\left|2x-3\right|=2x-3\) (1)

ĐKXĐ: \(\\ 2x-3\ge0\)

\(\Leftrightarrow x\ge\frac{3}{2}\)

\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}2x-3=2x-3\\2x-3=-2x+3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}0x=0\\4x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\forall x\in R\\x=\frac{3}{2}\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{\frac{3}{2}\right\}\)

a, Mình nghĩ là đề sai .

b, Ta có : \(\frac{x-45}{55}+\frac{x-47}{45}=\frac{x-55}{45}+\frac{x-53}{47}\)

=> \(\frac{x-45}{55}-1+\frac{x-47}{45}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)

=> \(\frac{x-45}{55}-\frac{55}{55}+\frac{x-47}{53}-\frac{53}{53}=\frac{x-55}{45}-\frac{45}{45}+\frac{x-53}{47}-\frac{47}{47}\)

=> \(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)

=> \(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)

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

=> \(x-100=0\)

=> \(x=100\)

Vậy phương trình trên có tập nghiệm là \(S=\left\{100\right\}\)

c, Ta có : \(\frac{2-x}{2010}-1=\frac{1-x}{2011}-\frac{x}{2012}\)

=> \(\frac{2-x}{2010}-1=\frac{1-x}{2011}+\frac{-x}{2012}\)

=> \(\frac{2-x}{2010}+1=\frac{1-x}{2011}+1+\frac{-x}{2012}+1\)

=> \(\frac{2-x}{2010}+\frac{2010}{2010}=\frac{1-x}{2011}+\frac{2011}{2011}+\frac{-x}{2012}+\frac{2012}{2012}\)

=> \(\frac{2012-x}{2010}=\frac{2012-x}{2011}+\frac{2012-x}{2012}\)

=> \(\frac{2012-x}{2010}-\frac{2012-x}{2011}-\frac{2012-x}{2012}=0\)

=> \(\left(2012-x\right)\left(\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)=0\)

=> \(2012-x=0\)

=> \(x=2012\)

Vậy phương trình trên có tập nghiệm là \(S=\left\{2012\right\}\)