Question

[nguyễn thị kim oanh]

có bao nhiêu cặp số nguyên x,y thõa mãn x.y-2x-3y=5

Answer 1

[Kiệt Nguyễn]

                            Giải

\(xy-2x-3y=5\)

\(\Leftrightarrow xy-3y-2x=5\)

\(\Leftrightarrow y\left(x-3\right)-2x+6=11\)

\(\Leftrightarrow y\left(x-3\right)-\left(2x-6\right)=11\)

\(\Leftrightarrow y\left(x-3\right)-2\left(x-3\right)=11\)

\(\Leftrightarrow\left(y-2\right)\left(x-3\right)=11\)

\(\Leftrightarrow\hept{\begin{cases}y-2\\x-3\end{cases}}\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)

Ta có bảng sau : 

\(x-3\)	\(-11\)	\(-1\)	\(1\)	\(11\)
\(y-2\)	\(-1\)	\(-11\)	\(11\)	\(1\)
\(x\)	\(-8\)	\(2\)	\(4\)	\(14\)
\(y\)	\(1\)	\(-9\)	\(13\)	\(3\)

Vậy có 4 cặp số nguyên x , y thỏa mãn \(\left(-8;1\right);\left(2;-9\right);\left(4;13\right);\left(14;3\right)\)

Answer 2

[Phạm Hồ Thanh Quang]

\(xy-2x-3y=5\Leftrightarrow x\left(y-2\right)-3\left(y-2\right)=11\Leftrightarrow\left(y-2\right)\left(x-3\right)=11\)\(\Leftrightarrow\hept{\begin{cases}x-3=1\\y-2=11\end{cases}}hay\hept{\begin{cases}x-3=11\\y-2=1\end{cases}}hay\hept{\begin{cases}x-3=-1\\y-2=-11\end{cases}}hay\hept{\begin{cases}x-3=-11\\y-2=-1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=4\\y=13\end{cases}}hay\hept{\begin{cases}x=14\\y=3\end{cases}}hay\hept{\begin{cases}x=2\\y=-9\end{cases}}hay\hept{\begin{cases}x=-8\\y=1\end{cases}}\)