Để \(\dfrac{2n}{n+3}+\dfrac{2}{n+3}=\dfrac{2n+2}{n+3}\) là số tự nhiên thì
\(\left\{{}\begin{matrix}2n+2⋮n+3\\\dfrac{2n+2}{n+3}>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2n+6-4⋮n+3\\\dfrac{n+1}{n+3}>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-4⋮n+3\\\left[{}\begin{matrix}n>=-1\\n< -3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n+3\in\left\{1;-1;2;-2;4;-4\right\}\\\left[{}\begin{matrix}n>=-1\\n< -3\end{matrix}\right.\end{matrix}\right.\)
=>\(n\in\left\{-4;-1;-5;1;-7\right\}\)
=>Có 5 số
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