Question

Tìm các số nguyên x và y sao cho
a) (x + 2)( y-1)=3
b) ( 3 -x)(xy +5) = -1

Answer 1

$a) (x + 2)(y - 1) = 3$

\(\cdot\left\{{}\begin{matrix}x+2=1\\y-1=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=4\end{matrix}\right.\left(TM\right)\)

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

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

\(\cdot\left\{{}\begin{matrix}x+2=-3\\y-1=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=0\end{matrix}\right.\left(TM\right)\)

Vậy ....

Answer 2

$b) (3 - x)(xy + 5) = - 1$

\(\cdot\left\{{}\begin{matrix}3-x=-1\\xy+5=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\4y+5=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\4y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-1\end{matrix}\right.\left(TM\right)\)

\(\cdot\left\{{}\begin{matrix}3-x=1\\xy+5=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y+5=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-3\end{matrix}\right.\left(TM\right)\)

Vậy $.................$