Question

necesito esto porfavor ​

Answer 1

Respuesta:

Explicación paso a paso:

$\textsf{Recordemos lo siguiente:}$

$\frac{a}{b}=\frac{c}{d}\\\\a*d=b*c\\\\\textsf{Aplicando lo anterior, a las expresiones,nos queda:}$

$0.\overline{6}=\frac{6}{9}=\frac{2}{3}\\\\\frac{\left(\frac{1}{2}+\frac{1}{4}\right)}{\sqrt[3]{\frac{1}{2}+2}}=\frac{x}{\frac{2}{3}}\\\\\frac{2}{3}*\left(\frac{1}{2}+\frac{1}{4}\right)=\sqrt[3]{\frac{1}{2}+2}*x\\\\\frac{2}{3}*\left(\frac{1}{2}+\frac{1}{4}\right)=\sqrt[3]{\frac{1+4}{2}}*x\\\\\frac{2}{3}\left(\frac{2+1}{4}\right)=\sqrt[3]{\frac{5}{2}}*x\\\\\frac{2}{3}\left(\frac{2}{4}\right)=\sqrt[3]{\frac{5}{2}}*x\\\\\frac{4}{12}=x*\sqrt[3]{\frac{5}{2}}\\\\x=\frac{\frac{4}{12}}{\sqrt[3]{\frac{5}{2}}}\\\\x=\frac{\frac{4}{12}}{\frac{\sqrt[3]{5}}{\sqrt[3]{2}}}\\\\x=\frac{4*\sqrt[3]{2}}{12*\sqrt[3]{5}}=\frac{4\sqrt[3]{2}}{12\sqrt[3]{5}}=\frac{1\sqrt[3]{2}}{3\sqrt[3]{5}}=\frac{1}{3}\sqrt[3]{\frac{2}{5}}$

$1,1\overline{5}=\frac{52}{45}\\\\1.4=\frac{14}{10}=\frac{7}{5}\\\\(1-1.4)^{2}=\left(1-\frac{7}{5}\right)=\left(\frac{-2}{5}\right)^{2}=\frac{4}{25}\\\\\left(-0.2+\frac{1}{2}+0.4\right)=\left(-0.6+\frac{1}{2}\right)=\left(-\frac{6}{10}+\frac{1}{2}\right)=\left(-\frac{3}{5}+\frac{1}{2}\right)=\\\\=-\frac{1}{10}\\\\\sqrt{1+\frac{5}{4}}=\sqrt{\frac{9}{4}}=\frac{3}{2}\\\texttt{Reuniendo todo y simplificando nos queda:}\\\\\frac{\left(1,1\overline{5}-\sqrt{1+\frac{5}{4}}\right)}{x}=\frac{(1-1,4)^{2}}{\left(-0.2+\frac{1}{2}-0.6\right)}\\\\\frac{\left(\frac{52}{45}-\frac{3}{2}\right)}{x}=\frac{\frac{4}{25}}{-\frac{1}{10}}\\\\-\frac{1}{10}*\left(\frac{52}{45}-\frac{3}{2}\right)=x*\frac{4}{25}\\\\x=\frac{-\frac{1}{10}*\left(\frac{52}{45}-\frac{3}{2}\right)}{\frac{4}{25}}$

$\sqrt{6,\overline{4}-1}=\sqrt{\frac{58}{9}-1}=\sqrt{\frac{49}{9}}=\frac{7}{3}\\\\(1,\overline{3}-1)=\left(\frac{4}{9}-1\right)=\left(\frac{2}{3}-1\right)=-\frac{1}{3}\\\\\left(\sqrt{6,\overline{4}-1}-(1,\overline{3}-1)\right)^{2}=\left(\frac{7}{3}-\left(-\frac{1}{3}\right)\right)^{2}=\left(\frac{7}{3}+\frac{1}{3}\right)^{2}=\left(\frac{8}{3}\right)^{2}=\\\\=\frac{64}{9}\\\\\frac{x}{\left(\sqrt[3]{0,64-1:25}\right)^{-1}}=x\sqrt[3]{\frac{64}{100}-1:25}=x\sqrt[3]{\frac{16}{25}-1:25}=x\sqrt[3]{\frac{16}{25}-\frac{1}{25}}=x\sqrt[3]{\frac{15}{25}}=\\\\=x*\sqrt[3]{\frac{3}{5}}\\\texttt{Reuniendo todo y simplificando nos queda:}\\\\\frac{\left[\sqrt{6,\overline{4}-1}-(1,\overline{3})\right]^{2}}{x}=\frac{x}{\left(\sqrt[3]{0,64}-1:25\right)^{-1}}\\\\\frac{\frac{64}{9}}{x}=x\sqrt[3]{\frac{3}{5}}\\\\\frac{64}{9}=x^{2}\sqrt[3]{\frac{3}{5}}\\\\x^{2}=\frac{\frac{64}{9}}{\sqrt[3]{\frac{3}{5}}}=\frac{64\sqrt[3]{5}}{9\sqrt[3]{3}}\\\\x=\pm\sqrt{\frac{64\sqrt[3]{5}}{9\sqrt[3]{3}}}$