Question

[Big City Boy]

Tìm các cặp số nguyên (x; y) thỏa mãn: x^2-xy+3x-y=5

Answer 1

[Trương Huy Hoàng]

x² - xy + 3x - y = 5

\(\Leftrightarrow\) x(x - y) + x - y + 2x = 5

\(\Leftrightarrow\) (x - y)(x + 1) + 2x + 2 = 7

\(\Leftrightarrow\) (x - y)(x + 1) + 2(x + 1) = 7

\(\Leftrightarrow\) (x - y + 2)(x + 1) = 7

Vì x, y \(\in\) Z nên (x - y + 2)(x + 1) \(\in\) Z

Xét các TH:

TH1: \(\left\{{}\begin{matrix}x-y+2=7\\x+1=1\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2-y=7\\x=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=0\\y=-5\end{matrix}\right.\) (TM)

TH2: \(\left\{{}\begin{matrix}x-y+2=-7\\x+1=-1\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}-2-y+2=-7\\x=-2\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\) (TM)

TH3: \(\left\{{}\begin{matrix}x-y+2=1\\x+1=7\end{matrix}\right.\)

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

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=6\\y=7\end{matrix}\right.\) (TM)

TH4: \(\left\{{}\begin{matrix}x-y+2=-1\\x+1=-7\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}-8-y+2=-1\\x=-8\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-8\\y=-5\end{matrix}\right.\) (TM)

Vậy ...

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