Question

[Ngọc Linh]

 

Answer 1

[Nguyễn Việt Lâm]

\(\Leftrightarrow x^2+2y^2+3xy-4x-5y+3=4\)

\(\Leftrightarrow\left(x^2+2xy-3x\right)+\left(xy+2y^2-3y\right)-\left(x+2y-3\right)=4\)

\(\Leftrightarrow x\left(x+2y-3\right)+y\left(x+2y-3\right)-\left(x+2y-3\right)=4\)

\(\Leftrightarrow\left(x+2y-3\right)\left(x+y-1\right)=4\)

Ta có bảng:

x+2y-3	-4	-2	-1	1	2	4
x+y-1	-1	-2	-4	4	2	1
x	1	-3	-8	6	1	-3
y	-1	2	5	-1	2	5

Vậy \(\left(x;y\right)=\left(1;-1\right);\left(-3;2\right);\left(-8;5\right);\left(6;-1\right);\left(1;2\right);\left(-3;5\right)\)