Question

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

Answer 1

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

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

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

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

\(\Leftrightarrow\) \(x=2003\) 

Answer 2

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

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

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

↔ x = 2003