Resolver ((23/9div3)^-2/94^2*frac{2/5})^-1

$$( \frac { ( 2 \frac { 3 } { 9 } \div 3 ) ^ { - 2 } } { \frac { 9 } { 4 } ^ { 2 } \cdot \frac { 2 } { 5 } } ) ^ { - 1 }$$

Calcular

$$\frac{49}{40}=1.225$$

Pasos da solución

Expresa \frac{\frac{2\times 9+3}{9}}{3} como unha única fracción.

$$\left(\frac{\left(\frac{2\times 9+3}{9\times 3}\right)^{-2}}{\left(\frac{9}{4}\right)^{2}\times \frac{2}{5}}\right)^{-1}$$

Multiplica 2 e 9 para obter 18.

$$\left(\frac{\left(\frac{18+3}{9\times 3}\right)^{-2}}{\left(\frac{9}{4}\right)^{2}\times \frac{2}{5}}\right)^{-1}$$

Suma 18 e 3 para obter 21.

$$\left(\frac{\left(\frac{21}{9\times 3}\right)^{-2}}{\left(\frac{9}{4}\right)^{2}\times \frac{2}{5}}\right)^{-1}$$

Multiplica 9 e 3 para obter 27.

$$\left(\frac{\left(\frac{21}{27}\right)^{-2}}{\left(\frac{9}{4}\right)^{2}\times \frac{2}{5}}\right)^{-1}$$

Reduce a fracción \frac{21}{27} a termos máis baixos extraendo e cancelando 3.

$$\left(\frac{\left(\frac{7}{9}\right)^{-2}}{\left(\frac{9}{4}\right)^{2}\times \frac{2}{5}}\right)^{-1}$$

Calcula \frac{7}{9} á potencia de -2 e obtén \frac{81}{49}.

$$\left(\frac{\frac{81}{49}}{\left(\frac{9}{4}\right)^{2}\times \frac{2}{5}}\right)^{-1}$$

Calcula \frac{9}{4} á potencia de 2 e obtén \frac{81}{16}.

$$\left(\frac{\frac{81}{49}}{\frac{81}{16}\times \frac{2}{5}}\right)^{-1}$$

Multiplica \frac{81}{16} e \frac{2}{5} para obter \frac{81}{40}.

$$\left(\frac{\frac{81}{49}}{\frac{81}{40}}\right)^{-1}$$

Divide \frac{81}{49} entre \frac{81}{40} mediante a multiplicación de \frac{81}{49} polo recíproco de \frac{81}{40}.

$$\left(\frac{81}{49}\times \frac{40}{81}\right)^{-1}$$

Multiplica \frac{81}{49} e \frac{40}{81} para obter \frac{40}{49}.

$$\left(\frac{40}{49}\right)^{-1}$$

Calcula \frac{40}{49} á potencia de -1 e obtén \frac{49}{40}.

$$\frac{49}{40}$$

Factorizar

$$\frac{7 ^ {2}}{2 ^ {3} \cdot 5} = 1\frac{9}{40} = 1.225$$