Question

\(\dfrac{x+1}{2006}+\dfrac{x+10}{1997}+\dfrac{x+19}{1988}=3\)

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Answer 1

Sửa đề :

\(\dfrac{x+1}{2006}+\dfrac{x+10}{1997}+\dfrac{x+19}{1988}=-3\)

\(\Leftrightarrow\left(\dfrac{x+1}{2006}+1\right)+\left(\dfrac{x+10}{1997}+1\right)+\left(\dfrac{x+19}{1988}+1\right)=0\)

\(\Leftrightarrow\dfrac{x+2007}{2006}+\dfrac{x+2007}{1997}+\dfrac{x+2007}{1988}=0\)

\(\Leftrightarrow\left(x+1007\right)\left(\dfrac{1}{2006}+\dfrac{1}{1997}+\dfrac{1}{1988}\right)=0\)

Mà \(\dfrac{1}{2006}+\dfrac{1}{1997}+\dfrac{1}{1988}\ne0\)

\(\Leftrightarrow x+2007=0\)

\(\Leftrightarrow x=-2007\)

Vậy..