Question

Tìm các số nguyên x;y thỏa x²+5y²+4xy+6x+12y+8=0

Answer 1

\(\Leftrightarrow x^2+4y^2+9+4xy+6x+12y+y^2-1=0\)

\(\Leftrightarrow\left(x+2y+3\right)^2=1-y^2\le1\)

TH1:\(\left\{{}\begin{matrix}\left(x+2y+3\right)^2=1\\y^2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=-4\\x=-2\end{matrix}\right.\\y=0\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}\left(x+2y+3\right)^2=0\\y^2=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=1\Rightarrow x=-5\\y=-1\Rightarrow x=-1\end{matrix}\right.\)

Vậy các cặp số nguyên t/m là \(\left(x;y\right)=\left(-4;0\right);\left(-2;0\right);\left(-5;1\right);\left(-1;-1\right)\)