a: (x-3)(y+2)=7
=>\(\left(x-3\right)\left(y+2\right)=1\cdot7=7\cdot1=\left(-1\right)\cdot\left(-7\right)=\left(-7\right)\cdot\left(-1\right)\)
=>\(\left(x-3;y+2\right)\in\left\{\left(1;7\right);\left(7;1\right);\left(-1;-7\right);\left(-7;-1\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(4;5\right);\left(10;-1\right);\left(2;-9\right);\left(-4;-3\right)\right\}\)
b: (x+4)(y-6)=11
=>\(\left(x+4\right)\left(y-6\right)=1\cdot11=11\cdot1=\left(-1\right)\cdot\left(-11\right)=\left(-11\right)\cdot\left(-1\right)\)
=>\(\left(x+4;y-6\right)\in\left\{\left(1;11\right);\left(11;1\right);\left(-1;-11\right);\left(-11;-1\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(-3;17\right);\left(7;7\right);\left(-5;-5\right);\left(-15;5\right)\right\}\)
c: ĐKXĐ: x<>3
Để y nguyên thì \(5⋮x-3\)
=>\(x-3\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{4;2;8;-2\right\}\)
d: ĐKXĐ: y<>-2
Để x nguyên thì \(3y+7⋮y+2\)
=>\(3y+6+1⋮y+2\)
=>\(1⋮y+2\)
=>\(y+2\in\left\{1;-1\right\}\)
=>\(y\in\left\{-1;-3\right\}\)
e: \(\left(x-1\right)^2+\left(y-2\right)^2=5\)
mà x,y nguyên
nên \(\left(x-1\right)^2+\left(y-2\right)^2=1+4=4+1\)
=>\(\left[\left(x-1\right)^2;\left(y-2\right)^2\right]\in\left\{\left(1;4\right);\left(4;1\right)\right\}\)
=>\(\left(x-1;y-2\right)\in\left\{\left(1;2\right);\left(-1;-2\right);\left(-2;-1\right);\left(2;1\right);\left(1;-2\right);\left(-2;1\right);\left(-1;2\right);\left(2;-1\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(2;4\right);\left(0;0\right);\left(-1;1\right);\left(3;3\right);\left(2;0\right);\left(-1;3\right);\left(0;4\right);\left(3;1\right)\right\}\)
