Упростите плииз, а то с корнями проблемы! Заранее спасибо

$(\frac{\sqrt{3}}{2-\sqrt{7-\sqrt{48}}}-\frac{2}{\sqrt{7+\sqrt{48}}-1})(2+\sqrt3)=\\\ \\\ =(\frac{\sqrt{3}}{2-\sqrt{4-4\sqrt{3}+3}}-\frac{2}{\sqrt{4+4\sqrt{3}+3}-1}})(2+\sqrt3)=\\\ \\\ =(\frac{\sqrt{3}}{2-\sqrt{2^2-2*2\sqrt{3}+(\sqrt{3})^2}}-\frac{2}{\sqrt{2^2+2*2\sqrt{3}+(\sqrt{3})^2}-1}})(2+\sqrt3)=\\\ \\\ =(\frac{\sqrt{3}}{2-\sqrt{(2-\sqrt{3})^2}}-\frac{2}{\sqrt{(2+\sqrt{3})^2}-1}})(2+\sqrt3)=\\\ \\\ = (\frac{\sqrt{3}}{2-|2-\sqrt{3}|}}-\frac{2}{|2+\sqrt{3}|-1}})(2+\sqrt3)$

так как 0" alt="2-\sqrt{3}>0" align="absmiddle" class="latex-formula"> то

$(\frac{\sqrt{3}}{2-(2-\sqrt{3})}}-\frac{2}{2+\sqrt{3}-1}})(2+\sqrt3)=(\frac{\sqrt{3}}{2-2+\sqrt{3}}}-\frac{2}{1+\sqrt{3}}})(2+\sqrt3)=\\\ \\\ =(\frac{\sqrt{3}}{\sqrt{3}}-\frac{2}{1+\sqrt{3}}})(2+\sqrt3)=(1-\frac{2}{1+\sqrt{3}}})(2+\sqrt3)=\\\ \\\ =\frac{1+\sqrt{3}-2}{1+\sqrt{3}}(2+\sqrt3)=\frac{\sqrt{3}-1}{1+\sqrt{3}}(2+\sqrt3)=\frac{(\sqrt{3}-1)(2+\sqrt3)}{1+\sqrt{3}}=\\\ \\\ =\frac{2\sqrt{3}+3-2-\sqrt3}{1+\sqrt{3}}=\frac{\sqrt{3}+1}{1+\sqrt{3}}=1$