0\; \; \Rightarrow \; \; (2-\sqrt3)^{x}=\frac{1}{t}>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\; \to \; \; (t-2)^2-3=0\; \to \\\\(t-2-\sqrt3)(t-2+\sqrt3)=0\; \; \Rightarrow \; \; t_1=2+\sqrt3\; ,\; \; t_2=2-\sqrt3\\\\a)\; (2+\sqrt3)^{x}=2+\sqrt3\; \; \to \; \; x=1" alt="(2+\sqrt3)^{x}+(2-\sqrt3)^{x}=4\\\\(2+\sqrt3)(2-\sqrt3)=2^2-(\sqrt3)^2=4-3=1\; \; \Rightarrow \; \; (2-\sqrt3)=\frac{1}{2+\sqrt3}\\ \\t=(2+\sqrt3)^{x}>0\; \; \Rightarrow \; \; (2-\sqrt3)^{x}=\frac{1}{t}>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\; \to \; \; (t-2)^2-3=0\; \to \\\\(t-2-\sqrt3)(t-2+\sqrt3)=0\; \; \Rightarrow \; \; t_1=2+\sqrt3\; ,\; \; t_2=2-\sqrt3\\\\a)\; (2+\sqrt3)^{x}=2+\sqrt3\; \; \to \; \; x=1" align="absmiddle" class="latex-formula">
