Question

tìm x;y nguyên biết:

Answer 1

\(\dfrac{1}{x}=\dfrac{1}{6}+\dfrac{y}{3}\)

\(\Rightarrow6=x\left(1+2y\right)\)

Vì \(x;y\in Z\Rightarrow x;1+2y\in Z\)

Mà \(1+2y\) là số lẻ `=>` \(x\) là số chẵn 

+) TH1:

\(\left\{{}\begin{matrix}x=6\\2y+1=1\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}x=6\\y=0\end{matrix}\right.\) 

+) TH2:

\(\left\{{}\begin{matrix}x=2\\2y+1=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

+) TH3:

\(\left\{{}\begin{matrix}x=-2\\2y+1=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-2\end{matrix}\right.\)

+) TH4:

\(\left\{{}\begin{matrix}x=-6\\2y+1=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=-1\end{matrix}\right.\)

Cả 4 TH đều thỏa mãn

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