Question

[Hoàng Anh Vũ]

tìm số nguyên x,y thỏa mãn: x^3-8y^3+7=0

Answer 1

[HT.Phong (9A5)]

\(x^3-8y^3+7=0\\ =>x^3-\left(2y\right)^3=-7\\ =>\left(x-2y\right)\left(x^2+2xy+4y^2\right)=-7\)

Vì x,y nguyên nên: 

TH1: \(\left\{{}\begin{matrix}x-2y=1\\x^2+2xy+4y^2=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2y=1\\\left(x-2y\right)^2+6xy=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2y=1\\xy=-\dfrac{4}{3}\end{matrix}\right.\) (loại vì x,y nguyên => xy nguyên)  

\(TH2:\left\{{}\begin{matrix}x-2y=7\\x^2+2xy+4y^2=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2y=7\\\left(x-2y\right)^2+6xy=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2y=7\\xy=-\dfrac{25}{3}\end{matrix}\right.\) (loại vì x,y nguyên => xy nguyên) 

TH3: \(\left\{{}\begin{matrix}x-2y=-1\\x^2+2xy+4y^2=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2y=-1\\\left(x-2y\right)^2+6xy=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2y=-1\\xy=1\end{matrix}\right.\) 

\(\Leftrightarrow\left\{{}\begin{matrix}x=2y-1\\\left(2y-1\right)y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y-1\\2y^2-y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y-1\\\left(y-1\right)\left(2y+1\right)=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\cdot1-1=1\\\left[{}\begin{matrix}y=1\left(tm\right)\\y=-\dfrac{1}{2}\left(ktm\right)\end{matrix}\right.\end{matrix}\right.\) 

TH4: \(\left\{{}\begin{matrix}x-2y=-7\\x^2+2xy+4y^2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2y=-7\\\left(x-2y\right)^2+6xy=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2y=-7\\xy=-8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2y-7\\\left(2y-7\right)y=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y-7\\2y^2-7y+8=0\end{matrix}\right.\Leftrightarrow x,y\in\varnothing\)

Vậy: x = 1 và y = 1