3.3.5. \( \left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{2}\right)^{-3} \) 3.3. Solve for \( x \) 3.3.1. \( 3 x=\frac{1}{9} \) 3.3.2. \( 2 x^{3}-1=53 \) 3.3.3. \( x^{2}=81 \)

To tackle \( \left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{2}\right)^{-3} \), remember that a negative exponent indicates the reciprocal. So, \( \left(\frac{1}{3}\right)^{-2} = 3^2 = 9 \) and \( \left(\frac{1}{2}\right)^{-3} = 2^3 = 8 \). Therefore, \( 9 + 8 = 17 \). Now, onto solving for \( x \)! 1. For \( 3x = \frac{1}{9} \), divide both sides by 3 to find \( x = \frac{1}{27} \). 2. In the equation \( 2x^3 - 1 = 53 \), add 1 to both sides to get \( 2x^3 = 54 \). Then divide by 2, yielding \( x^3 = 27 \). Take the cube root to find \( x = 3 \). 3. Finally, for \( x^2 = 81 \), taking the square root gives us \( x = 9 \) or \( x = -9 \)! Happy solving!