Đề sai sửa lại là:
\(x=\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\dfrac{125}{27}}}\)
\(\Leftrightarrow x=\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}+\sqrt[3]{3-\sqrt{9+\dfrac{125}{27}}}\)
\(\Leftrightarrow x^3=3+\sqrt{9+\dfrac{125}{27}}+3-\sqrt{9+\dfrac{125}{27}}+3.\left(\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}+\sqrt[3]{3-\sqrt{9+\dfrac{125}{27}}}\right)\left(\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}.\sqrt[3]{3-\sqrt{9+\dfrac{125}{27}}}\right)\)
\(\Leftrightarrow x^3=6+3x.\left(\dfrac{-5}{3}\right)\)
\(\Leftrightarrow x^3+5x-6=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow x=1\)
Vậy x là số nguyên
