Question

Tìm các số nguyên a,b biết: \(\frac{1}{a}+\frac{b}{6}=\frac{1}{3}\)với a≠0

Answer 1

\(\Leftrightarrow\frac{6+ab}{6a}=\frac{1}{3}\)

\(\Rightarrow18+3ab=6a\)

\(\Leftrightarrow\frac{3ab+18}{6}=3+\frac{ab}{2}=a\)

Để a nguyên thì ^{_{\(ab⋮2\Rightarrow ab=2k\left(k\in Z\right)\)}}

^{_{\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=2\\b=k\end{matrix}\right.\\\left\{{}\begin{matrix}a=k\\b=2\end{matrix}\right.\end{matrix}\right.\)}}

Vậy \(\left(a;b\right)=\left(2;k\right);\left(k;2\right)\)

Answer 2