Решить уравнение

$\tt \left(\sqrt{2+\sqrt{3}}\right)^x+\left(\sqrt{2-\sqrt{3}}\right)^x=4$

$\tt \left(\dfrac{1}{\sqrt{2-\sqrt{3}}}\right)^x+\left(\sqrt{2-\sqrt{3}}\right)^x=4\\ \\ \dfrac{1}{\left(\sqrt{2-\sqrt{3}}\right)^x}+\left(\sqrt{2-\sqrt{3}}\right)^x=4$

Пусть $\tt \left(\sqrt{2-\sqrt{3}}\right)^x=t$ , тогда получим

$\tt \dfrac{1}{t}+t=4~~~\bigg|\cdot t\ne 0\\ \\ t^2-4t+1=0\\ t^2-4t+4-3=0\\ (t-2)^2=3\\ \\ t=2\pm\sqrt{3}$

Возвращаемся к обратной замене

$\tt \left(\sqrt{2-\sqrt{3}}\right)^x=2-\sqrt{3}\\ \\ \left(2-\sqrt{3}\right)^{0.5x}=2-\sqrt{3}\\ \\ 0.5x=1\\ x=2$

$\tt \left(\sqrt{2-\sqrt{3}}\right)^x=2+\sqrt{3}\\ \\ \left(2-\sqrt{3}\right)^{0.5x}=(2-\sqrt{3})^{-1}\\ \\ 0.5x=-1\\ x=-2$

Ответ: ±2.

0)\\t^2-4t+1=0\\D/4=4-1=3\\t_1=2+\sqrt{3} \\ t_2 = 2-\sqrt{3}\\ \\ (\sqrt{2-\sqrt{3}})^x=t \\ \\ 1) (\sqrt{2-\sqrt{3}})^x = 2+\sqrt{3} |* (\sqrt{2+\sqrt{3}})^x\\ 1 = (\sqrt{2+\sqrt{3}})^{x+2}\\x+2=0\\x=-2\\\\2) (\sqrt{2-\sqrt{3}})^x= 2-\sqrt{3}\\x=2 " alt="(\sqrt{2+\sqrt{3}})^x+(\sqrt{2-\sqrt{3}})^x=4 |*(\sqrt{2-\sqrt{3}})^x\\1 + (\sqrt{2-\sqrt{3}})^{2x} =4(\sqrt{2-\sqrt{3}})^x\\(\sqrt{2-\sqrt{3}})^x=t (t>0)\\t^2-4t+1=0\\D/4=4-1=3\\t_1=2+\sqrt{3} \\ t_2 = 2-\sqrt{3}\\ \\ (\sqrt{2-\sqrt{3}})^x=t \\ \\ 1) (\sqrt{2-\sqrt{3}})^x = 2+\sqrt{3} |* (\sqrt{2+\sqrt{3}})^x\\ 1 = (\sqrt{2+\sqrt{3}})^{x+2}\\x+2=0\\x=-2\\\\2) (\sqrt{2-\sqrt{3}})^x= 2-\sqrt{3}\\x=2 " align="absmiddle" class="latex-formula">

Ответ: ±2