\(\left(\sqrt{2+\sqrt{3}}\right)^x+\left(\sqrt{2-\sqrt{3}}\right)^x=4\)
=>\(\left(\sqrt{2+\sqrt{3}}\right)^x+\dfrac{1}{\left(\sqrt{2+\sqrt{3}}\right)^x}=4\)(1)
Đặt \(\left(\sqrt{2+\sqrt{3}}\right)^x=t\)(t>=0)
(1) trở thành \(t+\dfrac{1}{t}=4\)
=>\(\dfrac{t^2+1-4t}{t}=0\)
=>\(t^2-4t+1=0\)
=>\(\left[{}\begin{matrix}t=2+\sqrt{3}\\t=2-\sqrt{3}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}\left(\sqrt{2+\sqrt{3}}\right)^x=2+\sqrt{3}\\\left(\sqrt{2+\sqrt{3}}\right)^x=2-\sqrt{3}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=2\\\left(2+\sqrt{3}\right)^{\dfrac{1}{2}x}=\left(2+\sqrt{3}\right)^{-1}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=2\\\dfrac{1}{2}x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
