Câu 1: Giải:
Gọi \(x=\dfrac{m}{n}\left(m,n\in Z;n\ne0;\left(m,n\right)=1\right)\) Khi đó:
\(x+\dfrac{1}{x}=\dfrac{m}{n}+\dfrac{n}{m}=\dfrac{m^2+n^2}{mn}\left(1\right)\)
Để \(x+\dfrac{1}{x}\in Z\Leftrightarrow m^2+n^2⋮mn\)
\(\Leftrightarrow m^2+n^2⋮n\Leftrightarrow n^2⋮m\Leftrightarrow n⋮m\)
Mà \(\left(m,n\right)=1\) \(\Leftrightarrow\left[{}\begin{matrix}m=1\\m=-1\end{matrix}\right.\)
\(*)\) Với \(m=1\) từ \(\left(1\right)\) ta có:
\(x+\dfrac{1}{x}=\dfrac{1^2+n^2}{1.n}=\dfrac{1+n^2}{n}.\)
Để \(x+\dfrac{1}{x}\in Z\Leftrightarrow1+n^2⋮n\Leftrightarrow1⋮n\) Hay \(n=\pm1\)
\(*)\) Với \(m=-1\) từ \(\left(1\right)\) ta có:
\(x+\dfrac{1}{x}=\dfrac{\left(-1\right)^2+n^2}{\left(-1\right).n}=\dfrac{1+n^2}{-n}\)
Để \(x+\dfrac{1}{x}\in Z\Leftrightarrow1+n^2⋮\left(-n\right)\Leftrightarrow1⋮\left(-1\right)\) Hay \(n=\pm1\)
Do đó \(x=\dfrac{m}{n}=\dfrac{1}{1}=\dfrac{1}{-1}=\dfrac{-1}{1}=\dfrac{-1}{-1}\) Hay \(x=\pm1\)
Vậy \(x=\pm1.\)