\(\dfrac{x+1}{2012}-\dfrac{x+2}{2011}=\dfrac{x+3}{2010}+1\)
\(\Leftrightarrow\dfrac{x+1}{2012}+1-\dfrac{x+2}{2011}-1=\dfrac{x+3}{2010}+1\)
\(\Leftrightarrow\dfrac{x+2013}{2012}-\dfrac{x+2013}{2011}-1=\dfrac{x+2013}{2010}\)
\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{2012}-\dfrac{1}{2011}-\dfrac{1}{2010}\right)=0\)
\(\Leftrightarrow\)\(x+2013=0\)(vì \(\dfrac{1}{2012}-\dfrac{1}{2011}-\dfrac{1}{2010}\ne0\))
\(\Leftrightarrow x=-2013\)
Vậy pt có tập nghiệm S=\(\left\{-2013\right\}\)
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