Question

Chứng minh rằng \(x_0=\sqrt{2+\sqrt{2+\sqrt{3}}}-\sqrt{6-3\sqrt{2+\sqrt{3}}}\) là 1 nghiệm của phương trình \(x^4-16x^2+32=0\)

Answer 1

\(x_0=\sqrt{2+\sqrt{2+\sqrt{3}}}-\sqrt{6-3\sqrt{2+\sqrt{3}}}\)(x₀>0)

=> \(\left(x_0\right)^2=2+\sqrt{2+\sqrt{3}}+6-3\sqrt{2+\sqrt{3}}-2\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{6-3\sqrt{2+\sqrt{3}}}\)

<=> \(\left(x_0\right)^2=8-2\sqrt{2+\sqrt{3}}-2\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right)\left(6-3\sqrt{2+\sqrt{3}}\right)}\)

<=> \(\left(x_0\right)^2=8-\sqrt{2}\sqrt{4+2\sqrt{3}}-2\sqrt{12-6\sqrt{2+\sqrt{3}}+6\sqrt{2+\sqrt{3}}-3\left(2+\sqrt{3}\right)}\)

<=> \(\left(x_0\right)^2=8-\sqrt{2}\sqrt{\left(\sqrt{3}+1\right)^2}-2\sqrt{12-6-3\sqrt{3}}=8-\sqrt{2}\left(\sqrt{3}+1\right)-2\sqrt{6-3\sqrt{3}}=8-\sqrt{2}\left(\sqrt{3}+1\right)-\sqrt{2}\sqrt{12-6\sqrt{3}}\)

<=> \(\left(x_0\right)^2=8-\sqrt{6}-\sqrt{2}-\sqrt{2}\sqrt{\left(3-\sqrt{3}\right)^2}=8-\sqrt{6}-\sqrt{2}-\sqrt{2}\left|3-\sqrt{3}\right|=8-\sqrt{6}-\sqrt{2}-\sqrt{2}\left(3-\sqrt{3}\right)\)

<=> \(\left(x_0\right)^2=8-\sqrt{6}-\sqrt{2}-3\sqrt{2}+\sqrt{6}=8-4\sqrt{2}\)

Có \(x^4-16x^2+32=0\) <=> \(x^4-8x^2+4\sqrt{2}x^2-8x^2+64-32\sqrt{2}-4\sqrt{2}x^2+32\sqrt{2}-32=0\)

<=> \(x^2\left(x^2-8+4\sqrt{2}\right)-8\left(x^2-8+4\sqrt{2}\right)-4\sqrt{2}\left(x^2-8+4\sqrt{2}\right)=0\)

<=>\(\left(x^2-8-4\sqrt{2}\right)\left(x^2-8+4\sqrt{2}\right)=0\)

=> \(\left[{}\begin{matrix}\left(x_1\right)^2=8+4\sqrt{2}\\\left(x_2\right)^2=8-4\sqrt{2}\end{matrix}\right.\) (x₁,x₂>0)

=> \(\left(x_0\right)^2=\left(x_2\right)^2\) <=> \(x_0=x_2\)( x₀,x₂>0)

Vậy x₀ là một nghiệm của pt \(x^4-16x^2+32=0\)