Ta có :\(x^2=2+\sqrt{2+\sqrt{3}}+6-3\sqrt{2+\sqrt{3}}-2\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right)\left(6-3\sqrt{2+\sqrt{3}}\right)}\)
\(=8-2\sqrt{2+\sqrt{3}}-2\sqrt{3\left(2+\sqrt{2+\sqrt{3}}\right)\left(2-\sqrt{2+\sqrt{3}}\right)}\)
\(=8-\frac{2}{\sqrt{2}}\sqrt{4+2\sqrt{3}}-2\sqrt{3\left(2^2-\sqrt{2+\sqrt{3}}^2\right)}\)
\(=8-\sqrt{2}\sqrt{\sqrt{3}^2+2\cdot1\sqrt{3}+1^2}-2\sqrt{3\left(4-2-\sqrt{3}\right)}\)
\(=8-\sqrt{2}\sqrt{\left(\sqrt{3}+1\right)^2}-2\sqrt{3}\sqrt{2-\sqrt{3}}\)
\(=8-\sqrt{2}\left(\sqrt{3}+1\right)-\frac{2\sqrt{3}}{\sqrt{2}}\sqrt{4-2\sqrt{3}}\)
\(=8-\left(\sqrt{6}+\sqrt{2}\right)-\sqrt{6}\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=8-\sqrt{6}-\sqrt{2}-\sqrt{6}\left(\sqrt{3}-1\right)\)
\(=8-\sqrt{6}-\sqrt{2}-\sqrt{18}+\sqrt{6}\)
\(=8-\sqrt{2}-\sqrt{18}\)
\(=8-\sqrt{2}\left(3+1\right)=8-4\sqrt{2}\)
\(\Rightarrow x^4-16x^2=\left(8-4\sqrt{2}\right)^2-16\left(8-4\sqrt{2}\right)\)
\(=8^2+4^2\cdot\sqrt{2}^2-2\cdot8\cdot4\sqrt{2}-16\cdot8+16\cdot4\sqrt{2}\)
\(=64+32-64\sqrt{2}-128+64\sqrt{2}\)
\(=-32\)
Vậy \(x^4-16x^2=-32\)
Tại hạ làm bừa có gì mong đạo hữu lượng thứ =))
