[tex]\frac{\sqrt{\sqrt3+2}}{2}=\frac{\sqrt{\frac{1}{2}(2\sqrt3+4)}}{2}=\frac{\sqrt{\frac{1}{2}(3+2\sqrt3+1)}}{2}=\frac{\sqrt{\frac{1}{2}(\sqrt3+1)^2}}{2}=\frac{{\frac{1}{\sqrt2}(\sqrt3+1)}}{2}=\frac{{\sqrt3+1}}{2\sqrt2}=\\\\=\frac{{\sqrt3+1}}{2\sqrt2}*\frac{\sqrt2}{\sqrt2}=\frac{\sqrt6+\sqrt2}{4}[/tex]
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[tex]\frac{\sqrt{\sqrt3+2}}{2}=\frac{\sqrt{\frac{1}{2}(2\sqrt3+4)}}{2}=\frac{\sqrt{\frac{1}{2}(3+2\sqrt3+1)}}{2}=\frac{\sqrt{\frac{1}{2}(\sqrt3+1)^2}}{2}=\frac{{\frac{1}{\sqrt2}(\sqrt3+1)}}{2}=\frac{{\sqrt3+1}}{2\sqrt2}=\\\\=\frac{{\sqrt3+1}}{2\sqrt2}*\frac{\sqrt2}{\sqrt2}=\frac{\sqrt6+\sqrt2}{4}[/tex]