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\(x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3+2\sqrt{2}}\)
Ta có: Đặt \(A=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\)=> \(A^2=\frac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{\sqrt{5}+1}\)
=> \(A^2=\frac{2\sqrt{5}+2\sqrt{5-4}}{\sqrt{5}+1}=\frac{2\left(\sqrt{5}+1\right)}{\sqrt{5}+1}=2\)=> \(A=\sqrt{2}\)
\(\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
==> \(x=\sqrt{2}-\left(\sqrt{2}+1\right)=-1\)
Do đó: N = (-1)2019 + 3.(-1)2020 - 2.(-1)2021 = -1 + 3 + 2 = 4
Ta có:
\(x=\frac{1}{2}.\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}=\frac{\sqrt{2}-1}{2}\)
\(\Rightarrow x\left(x+1\right)=\frac{\sqrt{2}-1}{2}.\frac{\sqrt{2}+1}{2}=\frac{1}{4}\)
Thế vô bài toán ta được
\(A=\left(4x^5+4x^4-5x^3+5x-2\right)^{2016}+2017\)
\(=\left(4x^4\left(x+1\right)-5x^3+5x-2\right)^{2016}+2017\)
\(=\left(-4x^3+5x-2\right)^{2016}+2017\)
\(=\left(\left(-4x^3-4x^2\right)+\left(4x^2+4x\right)+x-2\right)^{2016}+2017\)
\(=\left(-x+1+x-2\right)^{2016}+2017\)
\(=\left(-1\right)^{2016}+2017=2018\)
\(\frac{A}{\sqrt{2}}=\frac{1+\sqrt{7}}{2+\sqrt{8+2\sqrt{7}}}+\frac{1-\sqrt{7}}{2-\sqrt{8-2\sqrt{7}}}\)
\(=\frac{1+\sqrt{7}}{2+1+\sqrt{7}}+\frac{1-\sqrt{7}}{2-\sqrt{7}+1}\)
\(=\frac{1+\sqrt{7}}{3+\sqrt{7}}+\frac{1-\sqrt{7}}{3-\sqrt{7}}\)
=\(\frac{\left(1+\sqrt{7}\right)\left(3-\sqrt{7}\right)+\left(1-\sqrt{7}\right)\left(3+\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)
\(=\frac{-8}{2}=-4\)
\(\Rightarrow A=-4\sqrt{2}\)
Ta có:
\(x=\sqrt{3+\sqrt{5+2\sqrt{3}}}+\sqrt{3-\sqrt{5+2\sqrt{3}}}\) ( x> 0 )
\(\Rightarrow x^2=6+2\sqrt{\left(3+\sqrt{5+2\sqrt{3}}\right)\left(3-\sqrt{5+2\sqrt{3}}\right)}\)
\(=6+2\sqrt{9-5-2\sqrt{3}}\)
\(=6+2\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=6+2\sqrt{3}-2=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\)
\(\Rightarrow x=\sqrt{3}+1\)
Vậy :
\(A=x^2-2x-2=4+2\sqrt{3}-2\sqrt{3}-2-2\)
\(=0\)