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Giải:
\(\dfrac{2-x}{2007}-1=\dfrac{1-x}{2008}-\dfrac{x}{2009}\)
\(\Leftrightarrow\dfrac{2-x}{2007}-1+2=\dfrac{1-x}{2008}-\dfrac{x}{2009}+2\)
\(\Leftrightarrow\dfrac{2-x}{2007}+1=\dfrac{1-x}{2008}+1-\dfrac{x}{2009}+1\)
\(\Leftrightarrow\dfrac{2-x+2007}{2007}=\dfrac{1-x+2008}{2008}-\dfrac{x+2009}{2009}\)
\(\Leftrightarrow\dfrac{2009-x}{2007}=\dfrac{2009-x}{2008}-\dfrac{2009-x}{2009}\)
\(\Leftrightarrow\dfrac{2009-x}{2007}-\dfrac{2009-x}{2008}+\dfrac{2009-x}{2009}=0\)
\(\Leftrightarrow\left(2009-x\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}+\dfrac{1}{2009}\right)=0\)
Vì \(\dfrac{1}{2007}-\dfrac{1}{2008}+\dfrac{1}{2009}\ne0\)
\(\Leftrightarrow2009-x=0\)
\(\Leftrightarrow x=2009\)
Vậy ...
\(\dfrac{2-x}{2007}-1=\dfrac{1-x}{2008}-\dfrac{x}{2009}\)
\(\Leftrightarrow\left(\dfrac{2-x}{2007}+1\right)-\left(1+1\right)=\left(\dfrac{1-x}{2008}+1\right)-\left(\dfrac{x}{2009}+1\right)\)
\(\Leftrightarrow\dfrac{2-x+2007}{2007}=\dfrac{1-x+2008}{2008}-\dfrac{x+2009}{2009}\)
\(\Leftrightarrow\dfrac{2-x+2007}{2007}=\dfrac{1-x+2008}{2008}+\dfrac{-x+2009}{2009}\)
\(\Leftrightarrow\dfrac{2009-x}{2007}=\dfrac{2009-x}{2008}+\dfrac{2009-x}{2009}\)
\(\Leftrightarrow\left(2009-x\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2009}\right)=0\)
\(\Leftrightarrow2009-x=0\)
\(\Leftrightarrow x=2009\)