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Xét : \(1+2x=1+\frac{\sqrt{3}}{2}=\frac{2+\sqrt{3}}{2}=\frac{4+2\sqrt{3}}{4}=\frac{\left(\sqrt{3}+1\right)^2}{4}\)
\(1-2x=1-\frac{\sqrt{3}}{2}=\frac{2-\sqrt{3}}{2}=\frac{4-2\sqrt{3}}{4}=\frac{\left(\sqrt{3}-1\right)^2}{4}\)
Ta có : \(A=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{1+\sqrt{\left(\frac{\sqrt{3}+1}{2}\right)^2}}+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{1-\sqrt{\left(\frac{\sqrt{3}-1}{2}\right)^2}}\)
\(=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{1+\frac{\sqrt{3}+1}{2}}+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{1-\frac{\sqrt{3}-1}{2}}=\frac{\left(\sqrt{3}+1\right)^2}{2\left(3+\sqrt{3}\right)}+\frac{\left(\sqrt{3}-1\right)^2}{2\left(3-\sqrt{3}\right)}\)
\(=\frac{1}{2\sqrt{3}}\left(\frac{4+2\sqrt{3}}{\sqrt{3}+1}+\frac{4-2\sqrt{3}}{\sqrt{3}-1}\right)=\frac{1}{2\sqrt{3}}.\frac{4\sqrt{3}-4+6-2\sqrt{3}+4\sqrt{3}+4-6-2\sqrt{3}}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=\frac{1}{2\sqrt{3}}.\frac{4\sqrt{3}}{2}=1\)