a: xy+3x-y=6
=>x(y+3)-y-3=3
=>(x-1)(y+3)=3
=>\(\left(x-1;y+3\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(2;0\right);\left(4;-2\right);\left(0;-6\right);\left(-2;-4\right)\right\}\)
b: x+y-2xy=4
=>x-2xy+y=4
=>\(x\left(-2y+1\right)+y=4\)
=>\(-2x\left(y-0,5\right)+y-0,5=3,5\)
=>\(\left(-2x+1\right)\left(y-0,5\right)=3,5\)
=>\(\left(2x-1\right)\left(2y-1\right)=-7\)
=>\(\left(2x-1;2y-1\right)\in\left\{\left(1;-7\right);\left(-7;1\right);\left(-1;7\right);\left(7;-1\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(1;-3\right);\left(-3;1\right);\left(0;4\right);\left(4;0\right)\right\}\)
d: 2xy-x+y=6
=>\(x\left(2y-1\right)+y-0,5=5,5\)
=>\(2x\left(y-0,5\right)+\left(y-0,5\right)=5,5\)
=>\(\left(2x+1\right)\left(y-0,5\right)=5,5\)
=>\(\left(2x+1\right)\left(2y-1\right)=11\)
=>\(\left(2x+1;2y-1\right)\in\left\{\left(1;11\right);\left(11;1\right);\left(-1;-11\right);\left(-11;-1\right)\right\}\)
=>\(\left(2x;2y\right)\in\left\{\left(0;12\right);\left(10;2\right);\left(-2;-10\right);\left(-12;0\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(0;6\right);\left(5;1\right);\left(-1;-5\right);\left(-6;0\right)\right\}\)
a.Điều kiện x;y nguyên
xy+3x-y=6
=>x*(y+3)-y-3=6-3
=>(x-1)*(y+3)=3
Vì x y nguyên nên x-1 và y+3 nguyên
mà (x-1)*(y+3)=3
=>x-1 thuộc Ư (3)và y+3 thuộc Ư (3)
tự lập bảng tôi lười quá (a b c d giống nhau cách làm ý mà ) Đúng cho xin 1 like
