Đáp án: $ (x,y)\in\{(-\dfrac32, -2+2\sqrt{2}),(-\dfrac32, -2-2\sqrt{2}),(\dfrac{-3\pm2\sqrt{2}}{2},-2)\}$
Giải thích các bước giải:
Ta có:
$2xy+4x+3y+6=0$
$\to (2xy+4x)+(3y+6)=0$
$\to 2x(y+2)+3(y+2)=0$
$\to (2x+3)(y+2)=0$
$\to 2x+3=0$ hoặc $y+2=0$
$+)2x+3=0\to x=-\dfrac32$
Ta có:
$4x^2+y^2+12x+4y+5=0$
$\to (4x^2+12x+9)+(y^2+4y+4)-8=0$
$\to (2x+3)^2+(y+2)^2-8=0$
$\to 0^2+(y+2)^2-8=0$
$\to (y+2)^2=8$
$\to y+2=\pm2\sqrt{2}$
$\to y=-2\pm2\sqrt{2}$
$\to (x,y)\in\{(-\dfrac32, -2+2\sqrt{2}),(-\dfrac32, -2-2\sqrt{2})\}$
$+)y+2=0\to y=-2$
Mà $ (2x+3)^2+(y+2)^2-8=0$
$\to (2x+3)^2+0^2-8=0$
$\to (2x+3)^2=8$
$\to 2x+3=\pm2\sqrt{2}$
$\to x=\dfrac{-3\pm2\sqrt{2}}{2}$
$\to (x,y)=(\dfrac{-3\pm2\sqrt{2}}{2},-2)$
