Ответ б).
0\; \to \; |2-\sqrt3|=2-\sqrt3" alt="\sqrt{7-\sqrt{48}}=\sqrt{7-4\sqrt3}=\sqrt{2^2-2\cdot 2\cdot \sqrt3+(\sqrt3)^2}=\\\\=\sqrt{(2-\sqrt3)^2}=|2-\sqrt3|=2-\sqrt3\\\\P.S.\; \; 2-\sqrt3\approx 2-1,7>0\; \to \; |2-\sqrt3|=2-\sqrt3" align="absmiddle" class="latex-formula">