e) \( \left[\left(\frac{17}{25}\right)^{-5}\right]^{\frac{2}{5}} \div\left[\left(\frac{17}{25}\right)^{6}\right]^{-\frac{1}{3}} \)

We start with the expression \[ \left[\left(\frac{17}{25}\right)^{-5}\right]^{\frac{2}{5}} \div \left[\left(\frac{17}{25}\right)^{6}\right]^{-\frac{1}{3}}. \] **Step 1. Simplify the numerator** Apply the rule \(\left(a^b\right)^c = a^{bc}\): \[ \left[\left(\frac{17}{25}\right)^{-5}\right]^{\frac{2}{5}} = \left(\frac{17}{25}\right)^{-5 \cdot \frac{2}{5}} = \left(\frac{17}{25}\right)^{-2}. \] **Step 2. Simplify the denominator** Similarly, \[ \left[\left(\frac{17}{25}\right)^{6}\right]^{-\frac{1}{3}} = \left(\frac{17}{25}\right)^{6 \cdot \left(-\frac{1}{3}\right)} = \left(\frac{17}{25}\right)^{-2}. \] **Step 3. Divide the results** Now, divide the simplified numerator by the simplified denominator: \[ \frac{\left(\frac{17}{25}\right)^{-2}}{\left(\frac{17}{25}\right)^{-2}}. \] Using the property \(\frac{a^m}{a^n} = a^{m-n}\), we have: \[ \left(\frac{17}{25}\right)^{-2 - (-2)} = \left(\frac{17}{25}\right)^{0}. \] Since any nonzero number raised to the 0 power equals 1: \[ \left(\frac{17}{25}\right)^0 = 1. \] Thus, the final answer is \[ 1. \]