Question

tìm các số nguyên x, y thỏa mãn điều kiện  x(y+2)-y=3

Answer 1

  x (y+2) - y = 3

<=>( x-1 )y + 2y - 3 = 0

<=> x = 11 ;x=-2

Answer 2

Ta có: \(x.\left(y+2\right)-y-2=1\)            \(\left(x,y\inℤ\right)\)

    \(\Leftrightarrow x.\left(y+2\right)-\left(y+2\right)=1\)

    \(\Leftrightarrow\left(x-1\right).\left(y+2\right)=1=1.1=\left(-1\right).\left(-1\right)\)

+ \(\hept{\begin{cases}x-1=1\\y+2=1\end{cases}}\)\(\Rightarrow\)\(\hept{\begin{cases}x=2\left(TM\right)\\y=-1\left(TM\right)\end{cases}}\)

+ \(\hept{\begin{cases}x-1=-1\\y+2=-1\end{cases}}\)\(\Rightarrow\)\(\hept{\begin{cases}x=0\left(TM\right)\\y=-3\left(TM\right)\end{cases}}\)

Vậy ............

Answer 3

Ta có : \(x\left(y+2\right)-y=3\)

\(\Rightarrow x\left(y+2\right)-\left(y+2\right)+2=3\)

\(\Rightarrow\left(x-1\right)\left(y+2\right)=1\)

Vì \(x,y\inℤ\Rightarrow x-1;y+2\inℤ\)

\(\Rightarrow x-1;y+2\inƯ\left(1\right)=\){\(1;-1\)}

Ta có bảng sau : 

\(x-1\)	\(1\)	\(-1\)
\(y+2\)	\(1\)	\(-1\)
\(x\)	\(2\)	\(0\)
\(y\)	\(-1\)	\(-3\)

Vậy \(\left(x;y\right)\in\){\(\left(2;-1\right);\left(0;-3\right)\)}