\(=\left(x-y\right)\left(x+y\right)+7\left(x-y\right)=\left(x+y+7\right)\left(x-y\right)\)

C

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

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)=7\left(x-y\right)\)

\(\Leftrightarrow x^2+xy+y^2=7\)

Nếu \(y\ge2\Rightarrow x\ge3\Rightarrow x^2+xy+y^2>9>7\) (ktm)

\(\Rightarrow y< 2\Rightarrow y=1\)

\(\Rightarrow x^2+x+1=7\Rightarrow x\)

\(\Leftrightarrow x^3-y^3+7y-7x=0\\ \Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)-7\left(x-y\right)=0\\ \Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2-7\right)=0\\ \Leftrightarrow x^2+xy+y^2-7=0\left(x>y\Leftrightarrow x-y>0\right)\\ \Leftrightarrow x^2+xy+y^2=7\)

Vì \(x>y>0\) nên \(x^2< 7\)

Mà \(x\in Z\Leftrightarrow x^2\in\left\{1;4\right\}\)

Với \(x^2=1\Leftrightarrow\left[{}\begin{matrix}x=1\Rightarrow y^2+y-6=0\Rightarrow\left[{}\begin{matrix}y=2\\y=-3\end{matrix}\right.\\x=-1\Rightarrow y^2-y-6=0\Rightarrow\left[{}\begin{matrix}y=-2\\y=3\end{matrix}\right.\end{matrix}\right.\)

Với \(x^2=4\Leftrightarrow\left[{}\begin{matrix}x=2\Rightarrow y^2+2y-3=0\Rightarrow\left[{}\begin{matrix}y=-3\\y=1\end{matrix}\right.\\x=-2\Rightarrow y^2-2y-3=0\Rightarrow\left[{}\begin{matrix}y=-1\\y=3\end{matrix}\right.\end{matrix}\right.\)

Vậy ...(loại mấy TH x,y<0 ra)

Chọn đáp án B

⇒ Hệ phương trình đã cho vô nghiệm ⇒ S = ∅

Cách 2: Ta thấy:  ⇒ Hệ pt vô nghiệm

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

\(pt\Leftrightarrow x^3-y^3+7y-7x=0\)

\(\Leftrightarrow\left(x^3-y^3\right)-7\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)-7\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2-7\right)=0\)

xy(x - y) - 7(x - y) = (x - y)(xy - 7)