Решите уравнения пожалуйста!!! ПРОШУ

Решите задачу:

$1)\; \; (\sqrt{2+\sqrt3})^{x}+(\sqrt{2-\sqrt3})^{x}=4\\\\\\\sqrt{2+\sqrt3}= \frac{\sqrt{2+\sqrt3}\cdot \sqrt{2-\sqrt3}}{\sqrt{2-\sqrt3}}=\frac{\sqrt{4-3}}{\sqrt{2-\sqrt3}}= \frac{1}{\sqrt{2-\sqrt3}}\\\\\\t=(\sqrt{2+\sqrt3})^{x}\; \; \to \; \; (\sqrt{2-\sqrt3})^{x}=\frac{1}{t}\; ,\; \; t\ \textgreater \ 0\\\\t+\frac{1}{t}=4\\\\\frac{t^2-4t+1}{t}=0\; \; \to \; \; t^2-4t+1=0\; ,\; t\ne 0\\\\(t-2)^2-4+1=0\\\\(t-2)^2=3\; \; \to \; \; t-2=\pm \sqrt3\; ,\; \; t=2\pm \sqrt3\\\\a)\; \; (\sqrt{2+\sqrt3})^{x}=2+\sqrt3$

$(2+\sqrt3)^{\frac{x}{2}}=2+\sqrt3\\\\\frac{x}{2}=1\; \; \to \; \; x=2\\\\b)\; \; (\sqrt{2+\sqrt3})^{x}=2-\sqrt3\; ,\; \; \; \; \; \; 2-\sqrt3=\frac{1}{2+\sqrt3}\\\\(\sqrt{2+\sqrt3})^{x}=\frac{1}{2+\sqrt3}\; \; \to \; \; (\sqrt{2+\sqrt3})^{x}=(2+\sqrt3)^{-1}\\\\(2+\sqrt3)^{\frac{x}{2}}=(2+\sqrt3)^{-1}\\\\\frac{x}{2}=-1\; \; \to \; \; x=-2\\\\Otvet:\; \; x=-2,\; \; x=2\; .$

$2)\quad \sqrt{25-x^2}+\sqrt{15-x^2}=5\\\\\sqrt{25-x^2}-\sqrt{15-x^2}= \frac{(\sqrt{25-x^2}-\sqrt{15-x^2})(\sqrt{25-x^2}+\sqrt{15-x^2})}{\sqrt{25-x^2}+\sqrt{15-x^2}} =\\\\= \frac{(25-x^2)-(15-x^2)}{5} = \frac{10}{5}=2$