`\qquad 2yx^2-x+2y=3-2xy`
`<=>2yx^2+2xy+2y=x+3`
`<=>2y(x^2+x+1)=x+3`
`=>2y={x+3}/{x^2+x+1}`
Vì `y\in Z=>2y\in Z=>(x+3)\ \vdots \ (x^2+x+1)`
Vì `x\in Z=>x-2\in Z`
`=>(x-2)(x+3)\ \vdots \ (x^2+x+1)`
`=>(x^2+3x-2x-6)\ \vdots \ (x^2+x+1)`
`=>(x^2+x+1-7)\ \vdots \ (x^2+x+1)`
$\\$
Do `(x^2+x+1)\ \vdots \ (x^2+x+1)`
`=>7\ \vdots \ (x^2+x+1)`
`=>x^2+x+1\in Ư(7)={-7;-1;1;7}`
Ta có:
`x^2+x+1=x^2+2.x. 1/ 2 + 1/ 4 +3/ 4`
`=(x+ 1/ 2)^2+3/ 4\ge 3/ 4 >0` với mọi $x$
$\\$
`=>x^2+x+1\in {1;7}`
$\\$
+) Nếu `x^2+x+1=1`
`<=>x^2+x=0`
`<=>x(x+1)=0`
$⇔\left[\begin{array}{l}x=0\\x+1=0\end{array}\right.$ $⇔\left[\begin{array}{l}x=0\\x=-1\end{array}\right.$
$\\$
`\qquad 2y={x+3}/{x^2+x+1}`
`=>y={x+3}/{2(x^2+x+1)}`
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++) $TH: x=0$
`=>y={0+3}/{2.(0^2+0+1)}=3/ 2 ` (loại)
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++) $TH: x=-1$
`=>y={-1+3}/{2.[(-1)^2-1+1]}=1` (TM)
`=>(x;y)=(-1;1)`
$\\$
+) Nếu `x^2+x+1=7`
`<=>x^2+x-6=0`
`<=>x^2-2x+3x-6=0`
`<=>x(x-2)+3(x-2)=0`
`<=>(x-2)(x+3)=0`
$⇔\left[\begin{array}{l}x-2=0\\x+3=0\end{array}\right.$ $⇔\left[\begin{array}{l}x=2\\x=-3\end{array}\right.$
$\\$
`y={x+3}/{2(x^2+x+1)}`
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++) $TH: x=2$
`=>y={2+3}/{2.(2^2+2+1)}=5/ {14}` (loại)
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++) $TH: x=-3$
`=>y={-3+3}/{2.[(-3)^2-3+1]}=0` (TM)
`=>(x;y)=(-3;0)`
$\\$
Vậy nghiệm nguyên của phương trình là:
`(x;y)\in {(-1;1);(-3;0)}`
