Đặt \(B=\sqrt{2+\sqrt{\dfrac{5+\sqrt{5}}{2}}}+\sqrt{2-\sqrt{\dfrac{5+\sqrt{5}}{2}}}\)
=>\(B^2=2+\sqrt{\dfrac{5+\sqrt{5}}{2}}+2-\sqrt{\dfrac{5+\sqrt{5}}{2}}+2\cdot\sqrt{2^2-\left(\dfrac{5+\sqrt{5}}{2}\right)}\)
=>\(B^2=4+2\cdot\sqrt{4-\dfrac{5+\sqrt{5}}{2}}\)
=>\(B^2=4+2\cdot\sqrt{\dfrac{8-5-\sqrt{5}}{2}}\)
=>\(B^2=4+2\cdot\sqrt{\dfrac{3-\sqrt{5}}{2}}=4+2\cdot\sqrt{\dfrac{6-2\sqrt{5}}{4}}\)
=>\(B^2=4+2\cdot\dfrac{\left(\sqrt{5}-1\right)}{2}=4+\sqrt{5}-1=3+\sqrt{5}\)
=>\(B^2=\dfrac{6+2\sqrt{5}}{2}\)
=>\(B=\dfrac{\sqrt{6+2\sqrt{5}}}{\sqrt{2}}=\dfrac{\sqrt{5}+1}{\sqrt{2}}\)
\(x=\sqrt{2+\sqrt{\dfrac{5+\sqrt{5}}{2}}}+\sqrt{2-\sqrt{\dfrac{5+\sqrt{5}}{2}}}-\sqrt{3-\sqrt{5}}-1\)
\(=\dfrac{\sqrt{5}+1}{\sqrt{2}}-\sqrt{\dfrac{6-2\sqrt{5}}{2}}-1\)
\(=\dfrac{\sqrt{5}+1}{\sqrt{2}}-\dfrac{\sqrt{5}-1}{\sqrt{2}}-1=\dfrac{2}{\sqrt{2}}-1=\sqrt{2}-1\)
Thay \(x=\sqrt{2}-1\) vào A, ta được:
\(A=2\left(\sqrt{2}-1\right)^3+3\left(\sqrt{2}-1\right)^2-4\left(\sqrt{2}-1\right)+2\)
\(=2\left(-7+5\sqrt{2}\right)+3\left(3-2\sqrt{2}\right)-4\sqrt{2}+4+2\)
\(=-14+10\sqrt{2}+9-6\sqrt{2}-4\sqrt{2}+6\)
=-14+15
=1
