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\frac{\frac{\left(-\frac{1}{6}\right)^{2}}{\frac{5}{6}}-\sqrt{\frac{1}{9}}}{\sqrt[3]{\frac{1}{8}}+\left(1-\frac{1}{2}\right)^{2}\times \frac{9}{8}}
Resteu \frac{1}{2} de \frac{2}{3} per obtenir -\frac{1}{6}.
\frac{\frac{\frac{1}{36}}{\frac{5}{6}}-\sqrt{\frac{1}{9}}}{\sqrt[3]{\frac{1}{8}}+\left(1-\frac{1}{2}\right)^{2}\times \frac{9}{8}}
Calculeu -\frac{1}{6} elevat a 2 per obtenir \frac{1}{36}.
\frac{\frac{1}{36}\times \frac{6}{5}-\sqrt{\frac{1}{9}}}{\sqrt[3]{\frac{1}{8}}+\left(1-\frac{1}{2}\right)^{2}\times \frac{9}{8}}
Dividiu \frac{1}{36} per \frac{5}{6} multiplicant \frac{1}{36} pel recíproc de \frac{5}{6}.
\frac{\frac{1}{30}-\sqrt{\frac{1}{9}}}{\sqrt[3]{\frac{1}{8}}+\left(1-\frac{1}{2}\right)^{2}\times \frac{9}{8}}
Multipliqueu \frac{1}{36} per \frac{6}{5} per obtenir \frac{1}{30}.
\frac{\frac{1}{30}-\frac{1}{3}}{\sqrt[3]{\frac{1}{8}}+\left(1-\frac{1}{2}\right)^{2}\times \frac{9}{8}}
Torneu a escriure l'arrel quadrada de la divisió \frac{1}{9} com a divisió d'arrels quadrades \frac{\sqrt{1}}{\sqrt{9}}. Pren l'arrel quadrada del numerador i del denominador.
\frac{-\frac{3}{10}}{\sqrt[3]{\frac{1}{8}}+\left(1-\frac{1}{2}\right)^{2}\times \frac{9}{8}}
Resteu \frac{1}{30} de \frac{1}{3} per obtenir -\frac{3}{10}.
\frac{-\frac{3}{10}}{\frac{1}{2}+\left(1-\frac{1}{2}\right)^{2}\times \frac{9}{8}}
Calcula \sqrt[3]{\frac{1}{8}} i obté \frac{1}{2}.
\frac{-\frac{3}{10}}{\frac{1}{2}+\left(\frac{1}{2}\right)^{2}\times \frac{9}{8}}
Resteu 1 de \frac{1}{2} per obtenir \frac{1}{2}.
\frac{-\frac{3}{10}}{\frac{1}{2}+\frac{1}{4}\times \frac{9}{8}}
Calculeu \frac{1}{2} elevat a 2 per obtenir \frac{1}{4}.
\frac{-\frac{3}{10}}{\frac{1}{2}+\frac{9}{32}}
Multipliqueu \frac{1}{4} per \frac{9}{8} per obtenir \frac{9}{32}.
\frac{-\frac{3}{10}}{\frac{25}{32}}
Sumeu \frac{1}{2} més \frac{9}{32} per obtenir \frac{25}{32}.
-\frac{3}{10}\times \frac{32}{25}
Dividiu -\frac{3}{10} per \frac{25}{32} multiplicant -\frac{3}{10} pel recíproc de \frac{25}{32}.
-\frac{48}{125}
Multipliqueu -\frac{3}{10} per \frac{32}{25} per obtenir -\frac{48}{125}.