Đáp án+Giải thích các bước giải:
`\qquad |x+1|+|y+2|=3`
Vì `x;y\in ZZ=>|x+1|\in ZZ; |y+2|\in ZZ`
Với mọi `x;y\in ZZ=>|x+1|\ge 0; |y+2|\ge 0`
Do đó ta có các trường hợp sau:
+) $TH1:\begin{cases}|x+1|=0\\|y+2|=3\end{cases}$
`=>`$\left\{\begin{matrix}x+1=0\\\left[\begin{array}{l}y+2=3\\y+2=-3\end{array}\right.\end{matrix}\right.$`=>`$\left\{\begin{matrix}x=-1\\\left[\begin{array}{l}y=1\\y=-5\end{array}\right.\end{matrix}\right.$
`=>(x;y)\in {(-1;1);(-1;-5)}`
$\\$
+) $TH2:\begin{cases}|x+1|=1\\|y+2|=2\end{cases}$
`=>`$\left\{\begin{matrix}\left[\begin{array}{l}x+1=1\\x+1=-1\end{array}\right.\\\left[\begin{array}{l}y+2=2\\y+2=-2\end{array}\right.\end{matrix}\right.$`=>`$\left\{\begin{matrix}\left[\begin{array}{l}x=0\\x=-2\end{array}\right.\\\left[\begin{array}{l}y=0\\y=-4\end{array}\right.\end{matrix}\right.$
`=>(x;y)\in {(0;0);(0;-4);(-2;0);(-2;-4)}`
$\\$
+) $TH3:\begin{cases}|x+1|=2\\|y+2|=1\end{cases}$
`=>`$\left\{\begin{matrix}\left[\begin{array}{l}x+1=2\\x+1=-2\end{array}\right.\\\left[\begin{array}{l}y+2=1\\y+2=-1\end{array}\right.\end{matrix}\right.$`=>`$\left\{\begin{matrix}\left[\begin{array}{l}x=1\\x=-3\end{array}\right.\\\left[\begin{array}{l}y=-1\\y=-3\end{array}\right.\end{matrix}\right.$
`=>(x;y)\in {(1;-1);(1;-3);(-3;-1);(-3;-3)}`
$\\$
+) $TH4:\begin{cases}|x+1|=3\\|y+2|=0\end{cases}$
`=>`$\left\{\begin{matrix}\left[\begin{array}{l}x+1=3\\x+1=-3\end{array}\right.\\y+2=0\end{matrix}\right.$
`=>`$\left\{\begin{matrix}\left[\begin{array}{l}x=2\\x=-4\end{array}\right.\\y=-2\end{matrix}\right.$
`=>(x;y)\in {(2;-2);(-4;-2)}`
Vậy các cặp số nguyên `x;y` thỏa mãn đề bài là:
`(x;y)\in {(-1;1);(-1;-5);(0;0);(0;-4);(-2;0);(-2;-4);(1;-1);(1;-3);(-3;-1);(-3;-3);(2;-2);(-4;-2)}`
