Giải thích các bước giải:
Từ phương trình 1 ta suy ra :
$(x-y)(x^2+xy+y^2)+(x^2+y^2+xy)+(x-y+1)=0$
$\to (x-y+1)(x^2+xy+y^2)+(x-y+1)=0$
$\to (x-y+1)(x^2+xy+y^2+1)=0$
$\to x-y+1=0\to y=x+1$
$\to \sqrt{2x+x+1}+\sqrt{3x+2(x+1)+2}=3x^2-(x+1)+4$
$\to \sqrt{3x+1}+\sqrt{5x+4}=3x^2-x+3$
$\to (x+1-\sqrt{3x+1})+(x+2-\sqrt{5x+4})+3x^2-3x=0$
$\to \dfrac{(x+1-\sqrt{3x+1})(x+1+\sqrt{3x+1})}{x+1+\sqrt{3x+1}}+\dfrac{(x+2-\sqrt{5x+4})(x+2+\sqrt{5x+4})}{x+2+\sqrt{5x+4}}+3x(x-1)=0$
$\to \dfrac{x(x-1)}{x+1+\sqrt{3x+1}}+\dfrac{x(x-1)}{x+2+\sqrt{5x+4}}+3x(x-1)=0$
Vì $x\ge -\dfrac 13$
$\to x(x-1)=0\to x\in\{0,1\}\to y\in\{1,2\}$
