Đáp án: $ (x,y)\in\{(1,1), (-\dfrac45,-\dfrac{13}{5},(-2,1), (-2,-2)\}$
Giải thích các bước giải:
Ta có:
$3x^2+xy-4x+2y=2$
$\to xy+2y=-3x^2+4x+2$
$\to y(x+2)=-3x^2+4x+2$
Với $x=-2\to -2\cdot (-2+1)+y(y+1)=4$
$\to y^2+y-2=0$
$\to (y-1)(y+2)=0\to y\in\{1,-2\}$
$\to (x,y)\in\{(-2,1), (-2,-2)\}$
Với $x\ne -2$
$\to y=\dfrac{-3x^2+4x+2}{x+2}$
Thay vào phương trình $x(x+1)+y(y+1)=4$ ta được:
$x(x+1)+\dfrac{-3x^2+4x+2}{x+2}\cdot (\dfrac{-3x^2+4x+2}{x+2}+1)=4$
$\to x(x+1)+\dfrac{-3x^2+4x+2}{x+2}\cdot \dfrac{-3x^2+5x+4}{x+2}=4$
$\to x\left(x+1\right)+\dfrac{\left(-3x^2+5x+4\right)\left(-3x^2+4x+2\right)}{\left(x+2\right)^2}=4$
$\to x\left(x+1\right)\left(x+2\right)^2+\left(-3x^2+5x+4\right)\left(-3x^2+4x+2\right)=4\left(x+2\right)^2$
$\to 10x^4-22x^3+10x^2+30x+8=4x^2+16x+16$
$\to 10x^4-22x^3+6x^2+14x-8=0$
$\to 2\left(x-1\right)^3\left(5x+4\right)=0$
$\to x\in\{1,-\dfrac45\}$
$\to y\in\{1,-\dfrac{13}{5}\}$
$\to (x,y)\in\{(1,1), (-\dfrac45,-\dfrac{13}{5}\}$
