Ta có \(x=\dfrac{4\sqrt{4-\sqrt{5+\sqrt{21+\sqrt{80}}}}}{\sqrt{10}-\sqrt{2}}=\dfrac{4\sqrt{4-\sqrt{5+\sqrt{21+4\sqrt{5}}}}}{\sqrt{10}-\sqrt{2}}=\dfrac{4\sqrt{4-\sqrt{5+\sqrt{20+4\sqrt{5}+1}}}}{\sqrt{10}-\sqrt{2}}=\dfrac{4\sqrt{4-\sqrt{5+\sqrt{\left(2\sqrt{5}+1\right)^2}}}}{\sqrt{10}-\sqrt{2}}=\)=\(\dfrac{4\sqrt{4-\sqrt{5+2\sqrt{5}+1}}}{\sqrt{10}-\sqrt{2}}=\dfrac{4\sqrt{4-\sqrt{\left(\sqrt{5}+1\right)^2}}}{\sqrt{10}-\sqrt{2}}=\dfrac{4\sqrt{4-\sqrt{5}-1}}{\sqrt{10}-\sqrt{2}}=\dfrac{2\sqrt{2}\sqrt{6-2\sqrt{5}}}{\sqrt{10}-\sqrt{2}}=\dfrac{2\sqrt{2}\sqrt{5-2\sqrt{5}+1}}{\sqrt{10}-\sqrt{2}}=\dfrac{2\sqrt{2}\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{10}-\sqrt{2}}=\dfrac{2\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{10}-\sqrt{2}}=\dfrac{2\left(\sqrt{10}-\sqrt{2}\right)}{\sqrt{10}-\sqrt{2}}=2\)Vậy P=\(\left(x^3-4x+1\right)^{2018}=\left(2^3-4.2+1\right)^{2018}=\left(8-8+1\right)^{2018}=1^{2018}=1\)
