1. Ta có: x^3 + \sqrt{2} = 3(2\sqrt{2}-x^2)+3(2-x)
\Leftrightarrow x^3 + 3x^2+3x +1=5\sqrt{2}+7
\Leftrightarrow x^3 + 3x^2 + 3x + 1 = (\sqrt{2})^3+3 \times (\sqrt{2})^2 \times 1 + 3 \times \sqrt{2} \times 1^2 + 1^3
\Leftrightarrow (x+1)^3=(\sqrt{2}+1)^3
\Leftrightarrow x+1 = \sqrt{2} + 1 \Leftrightarrow x = \sqrt{2}
\star Với x = \sqrt{2}, ta có:
\color {red}{\mathbb {A} = \frac{(\sqrt{2}+1)^2}{2\sqrt{2}}-\sqrt{2}= 1 - \frac{\sqrt{2}}{4}}