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Question Number 173933 by mathlove last updated on 21/Jul/22

Commented by cortano1 last updated on 21/Jul/22

$$=1$$

Answered by som(math1967) last updated on 21/Jul/22

$$\frac{\frac{2+\sqrt{3}}{2}}{1+\sqrt{\frac{4+2\sqrt{3}}{4}}}+\frac{\frac{2-\sqrt{3}}{2}}{1-\sqrt{\frac{4-2\sqrt{3}}{4}}}$$$$=\frac{\frac{2+\sqrt{3}}{2}}{\frac{2+\sqrt{{(\sqrt{3}+1)}^2}}{2}}+\frac{\frac{2-\sqrt{3}}{2}}{\frac{2-\sqrt{{(\sqrt{3}-1)}^2}}{2}}$$$$=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}$$$$=\frac{(2+\sqrt{3})(3-\sqrt{3})+(2-\sqrt{3})(3+\sqrt{3})}{9-3}$$$$=\frac{6-2\sqrt{3}+3\sqrt{3}-3+6+2\sqrt{3}-3\sqrt{3}-3}{6}$$$$=\frac{12-6}{6}=1$$

Commented by mathlove last updated on 22/Jul/22

$$thanks$$

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