Giải thích các bước giải:
a.Ta có :
$x^2+y^2+1=2(xy-x+y)$
$\rightarrow x^2+y^2+1-2xy+2x-2y=0$
$\rightarrow (x-y+1)^2=0$
$\rightarrow x-y+1=0$
$\rightarrow y=x+1$
$\rightarrow x^3+3(x+1)^2+5x-12=(12-x-1)\sqrt{3-x}$
$\rightarrow x^3+3x^2+11x-9=(11-x)\sqrt{3-x}$
$\rightarrow x^3+3x^2+11x-9-11x+x^2=(11-x)(\sqrt{3-x}-x)$
$\rightarrow x^3+4x^2-9=(11-x)(\sqrt{3-x}-x)$
$\rightarrow (x+3)(x^2+x-3)=(11-x).\dfrac{3-x-x^2}{\sqrt{3-x}+x}$
$\rightarrow (x^2+x-3)(x+3-\dfrac{(11-x)}{\sqrt{3-x}+x})=0$
$\rightarrow (x^2+x-3)(x+3+\dfrac{x-11}{\sqrt{3-x}+x})=0$
$\rightarrow x^2+x-3=0\rightarrow x=\dfrac{-1-\sqrt{13}}{2},x\le 3$
$\rightarrow y=\dfrac{1-\sqrt{13}}{2}$
