Resolver {left(1/3right)}^2+{left(1/3right)}^-2+{left(-1/3right)}^3+3^-2

$${ \left( \frac{ 1 }{ 3 } \right) }^{ 2 } + { \left( \frac{ 1 }{ 3 } \right) }^{ -2 } + { \left(- \frac{ 1 }{ 3 } \right) }^{ 3 } + { 3 }^{ -2 }$$

Calcular

$$\frac{248}{27}\approx 9.185185185$$

Pasos da solución

Calcula \frac{1}{3} á potencia de 2 e obtén \frac{1}{9}.

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

Calcula \frac{1}{3} á potencia de -2 e obtén 9.

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

Suma \frac{1}{9} e 9 para obter \frac{82}{9}.

$$\frac{82}{9}+\left(-\frac{1}{3}\right)^{3}+3^{-2}$$

Calcula -\frac{1}{3} á potencia de 3 e obtén -\frac{1}{27}.

$$\frac{82}{9}-\frac{1}{27}+3^{-2}$$

Resta \frac{1}{27} de \frac{82}{9} para obter \frac{245}{27}.

$$\frac{245}{27}+3^{-2}$$

Calcula 3 á potencia de -2 e obtén \frac{1}{9}.

$$\frac{245}{27}+\frac{1}{9}$$

Suma \frac{245}{27} e \frac{1}{9} para obter \frac{248}{27}.

$$\frac{248}{27}$$

Factorizar

$$\frac{2 ^ {3} \cdot 31}{3 ^ {3}} = 9\frac{5}{27} = 9.185185185185185$$