\(\sqrt{x+2}-y^3=\sqrt{y+2}-x^3\)
\(\Leftrightarrow\left(\sqrt{x+2}-\sqrt{y+2}\right)+\left(x^3-y^3\right)=0\)
\(\Leftrightarrow\dfrac{x+2-y-2}{\sqrt{x+2}+\sqrt{y+2}}+\left(x-y\right)\left(x^2-xy+y^2\right)=0\)
\(\Leftrightarrow\left(\dfrac{1}{\sqrt{x+2}+\sqrt{y+2}}+x^2-xy+y^2\right)\left(x-y\right)=0\)
⇒ x = y. Thay vào A
\(\Rightarrow A=x^2+2x^2-2x^2+2x+10\)
\(=\left(x+1\right)^2+9\ge9\)
Suy ra Min A = 9 ⇔ x = y = - 1
\(A=x^2+2xy-2y^2+2y+10\)
\(\Leftrightarrow A=x^2+2xy+y^2-3y^2+2y-\dfrac{1}{3}+\dfrac{31}{3}\)
\(\Leftrightarrow A=\left(x^2+2xy+y^2\right)-\left(3y^2-2y+\dfrac{1}{3}\right)+\dfrac{31}{3}\)
\(\Leftrightarrow A=\left(x+y\right)^2-3\left(y^2-\dfrac{2}{3}y+\dfrac{1}{9}\right)+\dfrac{31}{3}\)
\(\Leftrightarrow A=\left(x+y\right)^2-3\left[y^2-2.y.\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2\right]+\dfrac{31}{3}\)
\(\Leftrightarrow A=\left(x+y\right)^2-3\left(y-\dfrac{1}{3}\right)^2+\dfrac{31}{3}\)
Vậy GTNN của \(A=\dfrac{31}{3}\) khi \(\left\{{}\begin{matrix}x+y=0\\y-\dfrac{1}{3}=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{3}=0\\y=\dfrac{1}{3}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)
