Giải phương trình : $\dfrac{2-x}{2001}-1= \dfrac{1-x}{2002}-\dfrac{x}{2003}$

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$\dfrac{2-x}{2001}-1= \dfrac{1-x}{2002}-\dfrac{x}{2003}$

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

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

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

Mà $\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}$ khác $0$

\Rightarrow $2003-x=0$ \Leftrightarrow $x=2003$
 
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