\(x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3+2\sqrt{2}}\)

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

\(x=\dfrac{\left(\sqrt{5}+2\right)\sqrt[3]{\left(\sqrt{5}-2\right)^3}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}=\dfrac{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}=\dfrac{5-4}{\sqrt{5}+3-\sqrt{5}}=\dfrac{1}{3}\)A=\(\left(3\left(\dfrac{1}{3}\right)^3+8\left(\dfrac{1}{3}\right)^2+2\right)^{2009}-3^{2009}=3^{2009}-3^{2009}=0\)

\(x=\dfrac{\sqrt[3]{17\sqrt{5}-38}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}.\left(\sqrt{5}+2\right)=\dfrac{\sqrt[3]{5\sqrt{5}-3.5.2+3\sqrt{5}.4-8}}{\sqrt{5}+\sqrt{9-2.3\sqrt{5}+5}}.\left(\sqrt{5}+2\right)=\dfrac{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}{3}=\dfrac{5-4}{3}=\dfrac{1}{3}\) Thay : \(x=\dfrac{1}{3}\) vào A , ta được :

\(A=\left(\dfrac{3}{27}+\dfrac{8}{9}-\dfrac{3}{3}+1\right)^{2012}=1^{2012}=1\)

Vậy ,...

ta có x=1 , thế vào f(x)

x=1/2

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 :

\(A=x^2-2x-2=4+2\sqrt{3}-2\sqrt{3}-2-2\)

\(=0\)

\(B=5\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}-\sqrt{\dfrac{5}{2}}\right)^2+2\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}-\sqrt{\dfrac{3}{2}}\right)^2\)\(B=5\left(2+\sqrt{3}+3-\sqrt{5}-\dfrac{5}{2}\right)+2\left(2-\sqrt{3}+3+\sqrt{5}-\dfrac{3}{2}\right)\)

\(B=10+5\sqrt{3}+15-5\sqrt{5}-\dfrac{25}{2}+4-2\sqrt{3}+6+2\sqrt{5}-3\)

\(B=\dfrac{39}{2}-3\sqrt{5}+3\sqrt{3}\)