a: 3xy-x+y=1
=>x(3y-1)+y-1/3=2/3
=>\(\left(y-\dfrac{1}{3}\right)\left(3x+1\right)=\dfrac{2}{3}\)
=>\(\left(3x+1\right)\left(3y-1\right)=2\)
=>\(\left(3x+1;3y-1\right)\in\left\{\left(1;2\right);\left(-1;-2\right);\left(2;1\right);\left(-2;-1\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(0;1\right);\left(-\dfrac{2}{3};-\dfrac{1}{3}\right);\left(\dfrac{1}{3};\dfrac{2}{3}\right);\left(-1;0\right)\right\}\)
mà (x;y) nguyên
nên \(\left(x;y\right)\in\left\{\left(0;1\right);\left(-1;0\right)\right\}\)
b: \(2x^2+3xy-2y^2=7\)
=>\(2x^2+4xy-xy-2y^2=7\)
=>\(2x\left(x+2y\right)-y\left(x+2y\right)=7\)
=>(x+2y)(2x-y)=7
=>\(\left(x+2y;2x-y\right)\in\left\{\left(1;7\right);\left(-1;-7\right);\left(7;1\right);\left(-7;-1\right)\right\}\)
TH1: \(\left\{{}\begin{matrix}x+2y=1\\2x-y=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x+4y=2\\2x-y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5y=-5\\2x-y=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-1\\2x=y+7=-1+7=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\left(nhận\right)\)
TH2: \(\left\{{}\begin{matrix}x+2y=-1\\2x-y=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x+4y=-2\\2x-y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5y=5\\x+2y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=1\\x=-1-2y=-1-2=-3\end{matrix}\right.\left(nhận\right)\)
TH3: \(\left\{{}\begin{matrix}x+2y=7\\2x-y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x+4y=14\\2x-y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5y=13\\x+2y=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{13}{5}\\x=7-2y=7-\dfrac{26}{5}=\dfrac{9}{5}\end{matrix}\right.\left(loại\right)\)
TH4: \(\left\{{}\begin{matrix}x+2y=-7\\2x-y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x+4y=-14\\2x-y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5y=-13\\2x-y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{13}{5}\\2x=y-1\end{matrix}\right.\left(loại\right)\)
Vậy: \(\left(x;y\right)\in\left\{\left(3;-1\right);\left(-3;1\right)\right\}\)
