Giải thích các bước giải:
$\sqrt{x+2}+x^3=\sqrt{y+2}+y^3$
$\to (x^3-y^3)+(\sqrt{x+2}-\sqrt{y+2})=0$
$\to (x-y)(x^2+xy+y^2)+\dfrac{x+2-(y+2)}{\sqrt{x+2}+\sqrt{y+2}}=0$
$\to (x-y)(x^2+xy+y^2)+\dfrac{x-y}{\sqrt{x+2}+\sqrt{y+2}}=0$
$\to (x-y)(x^2+xy+y^2+\dfrac{1}{\sqrt{x+2}+\sqrt{y+2}})=0$
$\to x-y=0$ vì $x^2+xy+y^2=(x+\dfrac 12y)^2+\dfrac 34y^2\ge 0 , x,y\ge -2$
$\to x=y$
$\to B=x^2+2x^2-2x^2+2x+10=x^2+2x+10=(x+1)^2+9\ge 9$
$\to Min B=9\to x=-1$
