Пожалуйста......

$(2+\sqrt3)^{x}+(2-\sqrt3)^{x}=14\\\\\\2+\sqrt3= \frac{(2+\sqrt3)(2-\sqrt3)}{2-\sqrt3} = \frac{4-3}{2-\sqrt3} =\frac{1}{2-\sqrt3}\\\\\\\frac{1}{(2-\sqrt3)^{x}}+(2-\sqrt3)^{x}=14\\\\a=(2-\sqrt3)^{x}\ \textgreater \ 0\; ;\; \; \; \frac{1}{a}+a-14=0\; ;\; \; \frac{a^2-14a+1}{a} =0\\\\a^2-14a+1=0\; ,\; D/4=49-1=48\; ,\\\\a_1=7-\sqrt{48}=7-4\sqrt3\; ,\; \; a_2=7+4\sqrt3\; ,\\\\(2-\sqrt3)^{x}=7-4\sqrt3=(2-\sqrt3)^2\; \; \to \; \; x=2\\\\(2-\sqrt3)^{x}=7+4\sqrt3=(2+\sqrt3)^2=\frac{1}{(2-\sqrt3)^2}=(2-\sqrt3)^{-2}\; \to \; x=-2$

Ответ: х=2, х=-2 .