Đặt x^4=a; 2x^2+1=b
Theo đề, ta có: a/b=2-b/a
=>\(\dfrac{a}{b}=\dfrac{2a-b}{a}\)
=>a^2=2ab-b^2
=>a^2-2ab+b^2=0
=>(a-b)^2=0
=>a=b
=>\(\dfrac{x^4}{2x^2+1}=\dfrac{2x^2+1}{x^4}\)
=>\(x^8=4x^4+4x^2+1\)
=>x^4=2x^2+1 hoặc x^4=-2x^2-1
=>\(\left[{}\begin{matrix}x^2=1+\sqrt{2}\\x^4+2x^2+1=0\left(loại\right)\end{matrix}\right.\Leftrightarrow x=\pm\sqrt{1+\sqrt{2}}\)