Lời giải:
Đặt \(x+\frac{\sqrt{2}+1}{2}=a\). Khi đó PT đã cho trở thành:
\((a+\frac{\sqrt{2}-1}{2})^4+(a-\frac{\sqrt{2}-1}{2})^4=33+12\sqrt{2}\)
\(\Leftrightarrow 2a^4+12a^2.(\frac{\sqrt{2}-1}{2})^2+2(\frac{\sqrt{2}-1}{2})^4=33+12\sqrt{2}\)
Coi đây là PT bậc 2 ẩn $a^2$.
\(\Delta'=36(\frac{\sqrt{2}-1}{2})^4-4(\frac{\sqrt{2}-1}{2})^4+2(33+12\sqrt{2})=100\)
\(\Rightarrow \left[\begin{matrix} a^2=\frac{-6(\frac{\sqrt{2}-1}{2})^2-10}{2}< 0(\text{loại})\\ a^2=\frac{-6(\frac{\sqrt{2}-1}{2})^2+10}{2}\end{matrix}\right.\)
Vậy \(a^2=\frac{-6(\frac{\sqrt{2}-1}{2})^2+10}{2}=\frac{11+6\sqrt{2}}{4}\)
\(\Rightarrow \left[\begin{matrix} a=\frac{3+\sqrt{2}}{2}\\ a=\frac{-3-\sqrt{2}}{2}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=1\\ x=-2-\sqrt{2}\end{matrix}\right.\)
Vậy....
