Bài 4:
\(x^3=9+4\sqrt{5}+9-4\sqrt{5}+3\sqrt[3]{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\\ \Leftrightarrow x^3=18+3x\sqrt[3]{81-80}=18+3x\\ \Leftrightarrow x^3-3x-18=0\left(đpcm\right)\)
Bài 5:
\(x^3=6+3\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\left(\sqrt[3]{3+2\sqrt{2}+\sqrt[3]{3-2\sqrt{2}}}\right)\\ \Leftrightarrow x^3=6+3x\\ \Leftrightarrow x^3-3x=6\\ y^3=34+3\sqrt[3]{\left(17+12\sqrt{2}\right)\left(17-12\sqrt{2}\right)}\left(\sqrt[3]{17+2\sqrt{2}}+\sqrt[3]{17-2\sqrt{2}}\right)\\ \Leftrightarrow y^3=34+3y\\ \Leftrightarrow y^3-3y=34\\ \Leftrightarrow P=x^3-3x+y^3-3y+2021=6+34+20212061\)
Bài 6: \(\left\{{}\begin{matrix}a=\sqrt{\left(3-2\sqrt{2}\right)^2}=3-2\sqrt{2}\\b=\sqrt{\left(3+2\sqrt{2}\right)^2}=3+2\sqrt{2}\end{matrix}\right.\)
\(a^2=17-12\sqrt{2}=18-12\sqrt{2}-1=6a-1\\ \Leftrightarrow a^2-6a+1=0\\ b^2=17+12\sqrt{2}=18+12\sqrt{2}-1=6b-1\\ \Leftrightarrow b^2-6b+1=0\)
