Вычислить √3+2/√3-2+4√7+4√3 1) 15 2) 15+8√3 3) 1 4) 1+8√3

[latex] \sqrt{3} + \frac{2}{ \sqrt{3}-2 }+4 \sqrt{7+4 \sqrt{3} }= \\ = \sqrt{3} + \frac{2}{ \sqrt{3}-2 }+4 \sqrt{2^{2}+2*2* \sqrt{3}+ (\sqrt{3})^{2} }= \\ = \sqrt{3} + \frac{2}{ \sqrt{3}-2 }+4 \sqrt{(2+ \sqrt{3})^{2} }= \\ = \sqrt{3} + \frac{2}{ \sqrt{3}-2 }+4 (2+ \sqrt{3})= \\ =\sqrt{3} + \frac{2+4 (2+ \sqrt{3})( \sqrt{3}-2)}{ \sqrt{3}-2 }= \\ =\sqrt{3} + \frac{2+4 ( (\sqrt{3})^{2}-2^{2})}{ \sqrt{3}-2 }= \\ =\sqrt{3} + \frac{2+4 (3-4)}{ \sqrt{3}-2 }= \\ [/latex] [latex]=\sqrt{3} + \frac{2-4}{ \sqrt{3}-2 }= \\ =\sqrt{3} + \frac{-2}{ \sqrt{3}-2 }= \\ = \frac{\sqrt{3}(\sqrt{3}-2) -2}{ \sqrt{3}-2 }= \\ = \frac{3-2 \sqrt{3}-2}{ \sqrt{3}-2 }= \\ = \frac{1-2 \sqrt{3}}{ \sqrt{3}-2 }= \\ [/latex] [latex]= \frac{(1-2 \sqrt{3}) (\sqrt{3}+2)}{ (\sqrt{3}-2 ) (\sqrt{3}+2) }= \\ = \frac{(1-2 \sqrt{3}) (\sqrt{3}+2)}{ (\sqrt{3})^{2}-2^{2} }= \\ = \frac{ \sqrt{3} +2-2*\sqrt{3}\sqrt{3}-4\sqrt{3}}{ (3-4) }= \\ = \frac{ -3\sqrt{3} +2-2*3}{ (3-4) }= \\ = \frac{ -3\sqrt{3} -4}{ (-1) }= \\ =4+3 \sqrt{3} \\ [/latex]