Câu 4:
[tex]x=\frac{1}{\sqrt[3]{4-\sqrt{15}}}+\sqrt[3]{4-\sqrt{15}}\\\Rightarrow x^3=\frac{1}{4-\sqrt{15}}+4-\sqrt{15}+3\frac{1}{\sqrt[3]{4-\sqrt{15}}}.\sqrt[3]{4-\sqrt{15}}\left ( \frac{1}{\sqrt[3]{4-\sqrt{15}}}+\sqrt[3]{4-\sqrt{15}} \right )\\\Leftrightarrow x^3=\frac{4+\sqrt{15}}{16-15}+4-\sqrt{15}+3x\\\Leftrightarrow x^3=8+3x\\\Leftrightarrow x^3-3x=8[/tex]
Khi đó: [tex]y=x^3-3x+2002=8+2002=2010[/tex]