\begin{cases}\sqrt{2x-y-1}+\sqrt{3y+1}=\sqrt{x}+\sqrt{x+2y} \\ x^{3}-3x+2= 2y^{3}-y^{2}\end{cases} $3(x^{2}-2) + \frac{4\sqrt{2}}{\sqrt{x^{2}-x+1}} > \sqrt{x}(\sqrt{x-1} + 3\sqrt{x^{2}-1})$
\begin{cases}\sqrt{2x-y-1}+\sqrt{3y+1}=\sqrt{x}+\sqrt{x+2y} \\ x^{3}-3x+2= 2y^{3}-y^{2}\end{cases}$3(x^{2}-2) + \frac{4\sqrt{2}}{\sqrt{x^{2}-x+1}} > \sqrt{x}(\sqrt{x-1} + 3\sqrt{x^{2}-1})$
\begin{cases}\sqrt{2x-y-1}+\sqrt{3y+1}=\sqrt{x}+\sqrt{x+2y} \\ x^{3}-3x+2= 2y^{3}-y^{2}\end{cases} $3(x^{2}-2) + \frac{4\sqrt{2}}{\sqrt{x^{2}-x+1}} > \sqrt{x}(\sqrt{x-1} + 3\sqrt{x^{2}-1})$
Bài 1. $\begin{cases}\sqrt{2x-y-1}+\sqrt{3y+1}=\sqrt{x}+\sqrt{x+2y} \\ x^{3}-3x+2= 2y^{3}-y^{2}\end{cases}
$Bài 2. $3(x^{2}-2) + \frac{4\sqrt{2}}{\sqrt{x^{2}-x+1}} > \sqrt{x}(\sqrt{x-1} + 3\sqrt{x^{2}-1})$