Question

numeros irracionales ayuda por fa

Answer 1

Explicación paso a paso:

$\texttt{Simplificando las ra\'ices nos queda:}\\\sqrt{48}-\sqrt{12}+\sqrt{3}\\\sqrt{48}\\\sqrt{(3)(16)}=\sqrt{3}\sqrt{16}=\\=4\sqrt{3}\\\\\sqrt{12}=\sqrt{(4)(3)}=\sqrt{4}\sqrt{3}=\\=2\sqrt{3}\\\sqrt{48}-\sqrt{12}+\sqrt{3}=4\sqrt{3}-2\sqrt{3}+\sqrt{3}=\\=5\sqrt{3}-2\sqrt{3}=\boxed{3\sqrt{3}}\\\\\sqrt[3]{81}-\sqrt[3]{24}\\\\\sqrt[3]{(3)(27)}=\sqrt[3]{3}\sqrt[3]{27}=\\=\sqrt[3]{3}\sqrt[3]{3^{3}}=3\sqrt[3]{3}\\\sqrt[3]{(3)(8)}=\sqrt[3]{3}\sqrt[3]{2^{3}}=\\=2\sqrt[3]{3}\\\\\sqrt[3]{81}-\sqrt[3]{24}=\boxed{3\sqrt[3]{3}-2\sqrt[3]{3}=\sqrt[3]{3}}\\\\\sqrt{28}-\sqrt{7}+\sqrt{63}\\\sqrt{28}=\sqrt{(4)(7)}=\sqrt{4}\sqrt{7}=\\=2\sqrt{7}\\\sqrt{63}=\sqrt{(9)(7)}=\sqrt{9}\sqrt{7}=\\=3\sqrt{7}\\\\\sqrt{28}-\sqrt{7}+\sqrt{63}=2\sqrt{7}-\sqrt{7}+3\sqrt{7}=\\=2\sqrt{7}+3\sqrt{7}-\sqrt{7}=5\sqrt{7}-\sqrt{7}=\\=\boxed{4\sqrt{7}}$