Lời giải:

PT $\Leftrightarrow x^3+x+1-y(x^2-3)=0$

$\Leftrightarrow y=\frac{x^3+x+1}{x^2-3}$ (hiển nhiên $x^2-3\neq 0$ với mọi $x$ nguyên) 

Để $y$ nguyên thì $\frac{x^3+x+1}{x^2-3}$ nguyên 

$\Leftrightarrow x^3+x+1\vdots x^2-3$
$\Rightarrow x(x^2-3)+4x+1\vdots x^2-3$
$\Rightarrow 4x+1\vdots x^2-3$

Hiển nhiên $4x+1\neq 0$ nên $|4x+1|\geq x^2-3$
Nếu $x\geq \frac{-1}{4}$ thì $4x+1\geq x^2-3$
$\Leftrightarrow x^2-4x-4\leq 0$

$\Leftrightarrow (x-2)^2\leq 8<9$

$\Rightarrow -3< x-2< 3$

$\Rightarrow -1< x< 5$

$\Rightarrow x\in \left\{0; 1; 2; 3; 4\right\}$.

Nếu $x< \frac{-1}{4}$ thì $-4x-1\geq x^2-3$

$\Leftrightarrow x^2+4x-2\leq 0$

$\Leftrightarrow (x+2)^2-6\leq 0$

$\Leftrightarrow (x+2)^2\leq 6< 9$

$\Rightarrow -3< x+2< 3$
$\Rightarrow -5< x< 1$

$\Rightarrow x\in\left\{-4; -3; -2; -1\right\}$

Đến đây bạn thay vào tìm $y$ thôi

\(x^3+y^3-9xy=0\)

\(\Leftrightarrow\left(x+y\right)^3-3x^2y-3xy^2-9xy=0\)

\(\Leftrightarrow\left(x+y\right)^3+27-3xy\left(x+y+3\right)=27\)

\(\Leftrightarrow\left(x+y+3\right)\left[\left(x+y\right)^2-3\left(x+y\right)+9\right]-3xy\left(x+y+3\right)-27=0\)

\(\Leftrightarrow\left(x+y+3\right)\left(x^2+2xy+y^2-3x-3y+9-3xy\right)-27=0\)

\(\Leftrightarrow\left(x+y+3\right)\left(x^2-xy+y^2-3x-3y+9\right)-27=0\)

\(\Leftrightarrow\left(x+y+3\right)\left(2x^2-2xy+2y^2-6x-6y+18\right)-54=0\)

\(\Leftrightarrow\left(x+y+3\right)\left[\left(x-y\right)^2+\left(x-3\right)^2+\left(y-3\right)^2\right]=54\)

Do x, y > 0 => x + y + 3 > 3

Mà x, y nguyên dương => \(\left\{{}\begin{matrix}x+y+3\in Z^+\\\left(x-y\right)^2+\left(x-3\right)^2+\left(y-3\right)^2\in Z^+\end{matrix}\right.\)

Và \(\left(x-y\right)^2+\left(x-3\right)^2+\left(y-3\right)^2⋮2\)

TH1: \(\left\{{}\begin{matrix}x+y+3=9\\\left(x-y\right)^2+\left(x-3\right)^2+\left(y-3\right)^2=6\end{matrix}\right.\)

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

\(\Leftrightarrow x^2-x\left(6-x\right)+\left(6-x\right)^2-3x-3\left(6-x\right)=-6\)

\(\Leftrightarrow x^2-6x+8=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\Leftrightarrow y=2\left(tm\right)\\x=2\left(tm\right)\Leftrightarrow y=4\left(tm\right)\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x+y+3=27\\\left(x-y\right)^2+\left(x-3\right)^2+\left(y-3\right)^2=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=24\\x^2-xy+y^2-3x-3y=-8\end{matrix}\right.\)

\(\Leftrightarrow x^2-x\left(24-x\right)+\left(24-x\right)^2-3x-3\left(24-x\right)=-8\)

\(\Leftrightarrow3x^2-72x+512=0\) (vô nghiệm)

KL: Vậy phương trình có tập nghiệm (x;y) = [(2;4);(4;2)]

Ta có:

\(x^3+y^3-xy=7\)

\(\left(x+y\right)^3-3xy\left(x+y\right)-xy=7\)

Thay x+y = 3 ta dc:

\(3^3-9xy-xy=7\)

\(-10xy=-20\)

\(xy=2\)

Vậy, tập hợp x, y thoả mãn đaẻng thức là: {x,y thuộc R/xy=2}

Đáp án đúng : B