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\(\Leftrightarrow x^2+4y^2+9+4xy+6x+12y+y^2-1=0\)
\(\Leftrightarrow\left(x+2y+3\right)^2=1-y^2\le1\)
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 \(\left(x;y\right)=\left(-4;0\right);\left(-2;0\right);\left(-5;1\right);\left(-1;-1\right)\)
\(a,x^2-4xy+5y^2=169\\ \Leftrightarrow\left(x-2y\right)^2+y^2=169\\ Vìx,y\in Znên:\\ \left[{}\begin{matrix}\left\{{}\begin{matrix}\left(x-2y\right)^2=0\\y^2=169\end{matrix}\right.\\\left\{{}\begin{matrix}\left(x-2y\right)^2=169\\y^2=0\end{matrix}\right.\\\left\{{}\begin{matrix}\left(x-2y\right)^2=25\\y^2=144\end{matrix}\right.\\\left\{{}\begin{matrix}\left(x-2y\right)^2=144\\y^2=25\end{matrix}\right.\end{matrix}\right.\\ Giảira\)
a)
\(\Leftrightarrow yz=z^2+2z+3\Leftrightarrow z\left(y-2-z\right)=3\)
\(\hept{\begin{cases}z=\left\{-3,-1,1,3\right\}\\y-2-z=\left\{-1,-3,3,1\right\}\end{cases}\Rightarrow\hept{\begin{cases}x=\left\{-2,0,2,4\right\}\\y=\left\{-2,-4,6,6\right\}\end{cases}}}\)