Question

Giải phương trình nghiệm nguyên: x²+3y²+4xy-2x-6y=5

Answer 1

\(\Leftrightarrow x^2+3xy+3y^2+xy-2x-6y=5\)

\(\Leftrightarrow x\left(x+3y\right)+y\left(x+3y\right)-2\left(x+3y\right)=5\)

\(\Leftrightarrow\left(x+y-2\right)\left(x+3y\right)=5\)

Bảng giá trị:

x+y-2	-5	-1	1	5
x+3y	-1	-5	5	1
x	-4	4	2	10
y	1	-3	1	-3

Vậy \(\left(x;y\right)=\left(-4;1\right);\left(4;-3\right);\left(2;1\right);\left(10;-3\right)\)