Question

Ayuda con procedimiento
doy coronita​

Answer 1

Respuesta:

$\left[\left(\frac{8}{10}\right)^{2}-\left(\frac{4}{10}\right)^{2}+\sqrt{\left(\frac{7}{3}\right)^{2}}=\frac{211}{75}$

$\left(sqrt{\frac{64}{216}}\right)\times\left(\sqrt{\frac{64}{216}}\right)-\sqrt{\frac{36}{100}}=-\frac{41}{135}$

Explicación paso a paso:

Recordemos que:

$\left(\sqrt[n]{a}\right)^{n}=\sqrt[n]{a^{n}}=a$

$\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$

Aplicando las propiedades anteriores nos queda:

$a).-\\\\\\sqrt{\left(\frac{7}{3}\right)^{2}}=\frac{7}{3}\\\\\left[\left(\frac{8}{10}\right)^{2}-\left(\frac{4}{10}\right)^{2}\right]+\sqrt{\left(\frac{7}{3}\right)^{2}}=\\\\=\left[\left(\frac{64}{100}\right)-\left(\frac{16}{100}\right)\right]+\frac{7}{3}$

$\frac{64}{100}-\frac{16}{100}+\frac{7}{3}=\frac{48}{100}=\frac{12}{25}\\\\\frac{12}{25}+\frac{7}{3}=\frac{211}{75}$

$\left(\sqrt{\frac{64}{216}}\right)\times\left(\sqrt{\frac{64}{216}}\right)-\sqrt{\frac{36}{100}}$

$\left(\frac{\sqrt{64}}{\sqrt{216}}\right)\times\left(\frac{\sqrt{64}}{\sqrt{216}}\right)-\frac{\sqrt{36}}{\sqrt{100}}=\\\\=\left(\frac{8}{6\sqrt{6}}\right)\times\left(\frac{8}{6\sqrt{6}}\right)-\frac{6}{10}=\left(\frac{8\times8}{6^{2}(\sqrt{6})^{2}}\right)-\frac{6}{10}$

$\frac{64}{216}-\frac{6}{10}=-\frac{41}{135}$