Đáp án:
`(x;y)\in {(1;1);(3;3);(3;2);(1;2)}`
Giải thích các bước giải:
`\qquad x^2+2y^2+2<2xy+4y`
`<=>(x^2-2xy+y^2)+(y^2-4y+4)-2<0`
`<=>(x-y)^2+(y-2)^2<2`
Với mọi `x;y` ta có: $\begin{cases}(x-y)^2\ge 0\\(y-2)^2\ge 0\end{cases}$
Vì `x;y\in ZZ=>(x-y)^2\in ZZ; (y-2)^2\in ZZ`
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+) $TH1: \begin{cases}(x-y)^2=0\\(y-2)^2=1\end{cases}$
`<=>`$\left\{\begin{matrix}x=y\\\left[\begin{array}{l}y-2=1\\y-2=-1\end{array}\right.\end{matrix}\right.$`<=>`$\left\{\begin{matrix}x=y\\\left[\begin{array}{l}y=3\\y=1\end{array}\right.\end{matrix}\right.$
`=>(x;y)\in {(1;1);(3;3)}`
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+) $TH2: \begin{cases}(y-2)^2=0\\(x-y)^2=1\end{cases}$
`<=>`$\left\{\begin{matrix}y=2\\\left[\begin{array}{l}x-y=1\\x-y=-1\end{array}\right.\end{matrix}\right.$
`<=>`$\left\{\begin{matrix}y=2\\\left[\begin{array}{l}x=y+1=2+1=3\\x=y-1=2-1=1\end{array}\right.\end{matrix}\right.$
`=>(x;y)\in {(3;2);(1;2)}`
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Vậy `(x;y)\in {(1;1);(3;3);(3;2);(1;2)}` thỏa mãn đề bài
