Question

. Transformar a radical simple:
7-√48

Answer 1

Explocación paso a paso:

$\sqrt{7-\sqrt{48}}\\\texttt{Al simplificar el radical}\,\sqrt{48}\,\texttt{nos queda:}\\\sqrt{48}=\sqrt{(3)(16)}=\\=\sqrt{3}\sqrt{16}=4\sqrt{3}\\\texttt{Luego aplicando:}\,A=7\,\sqrt{B}=4\sqrt{3}\,\texttt{y calculando}\,C\,\texttt{se obtiene}\\C=\sqrt{A^{2}-(\sqrt{B})^{2}}=\sqrt{(7)^{2}-(4\sqrt{3})^{2}}=\\=\sqrt{49-48}=\sqrt{1}=1\\\texttt{Luego}\,\sqrt{7-\sqrt{48}}=\sqrt{\frac{A+C}{2}}-\sqrt{\frac{A-C}{2}}=\sqrt{\frac{7+1}{2}}-\sqrt{\frac{7-1}{2}}=\\=\sqrt{\frac{8}{2}}-\sqrt{\frac{6}{2}}=\sqrt{4}-\sqrt{3}=\\=\boxed{2-\sqrt{3}}$