a x.y = x + y + 1992

⇔ x.y - x - y = 1992

⇔ x(y - 1) - y + 1 = 1993

⇔ x(y - 1) - (y - 1) = 1993

⇔ (y - 1)(x - 1) = 1993

TH1: \(\left\{{}\begin{matrix}y-1=1\\x-1=1993\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=1994\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}y-1=1993\\x-1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1994\\x=2\end{matrix}\right.\)

TH3: \(\left\{{}\begin{matrix}y-1=-1\\x-1=-1993\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=-1992\end{matrix}\right.\)

TH4: \(\left\{{}\begin{matrix}y-1=-1993\\x-1=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1992\\x=0\end{matrix}\right.\)

Vậy cặp số (x;y) thỏa mãn là: (1994;2); (2;1994); (-1992;0); (0; -1992)

xy+ x - 2y = 5

=> x(y+1) - 2y = 5

=> x(y+1) - 2y - 2 = 3

=> x(y+1) - 2 ( y+1) = 3

=> (x-2)(y+1)= 3

 vay (x,y)= (1;-4);(3;2);(-1;-2);(5;0)

cuc cac

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