19. \( \frac{2^{x+}}{\left(2^{x}\right)^{x-1}} \div \frac{(4)^{x+1}}{\left(2^{x-1}\right)^{x+1}} \)

To simplify the expression \( \frac{2^{x+}}{\left(2^{x}\right)^{x-1}} \div \frac{(4)^{x+1}}{\left(2^{x-1}\right)^{x+1}} \), we will break it down step by step. 1. **Convert all terms to base 2**: - We know that \( 4 = 2^2 \), so \( (4)^{x+1} = (2^2)^{x+1} = 2^{2(x+1)} = 2^{2x + 2} \). - The expression simplifies to: \[ \frac{2^{x+}}{\left(2^{x}\right)^{x-1}} \div \frac{2^{2x + 2}}{\left(2^{x-1}\right)^{x+1}}. \] 2. **Simplify each component**: - The denominator \( \left(2^{x}\right)^{x-1} = 2^{(x)(x-1)} = 2^{x^2 - x} \). - The denominator in the second fraction \( \left(2^{x-1}\right)^{x+1} = 2^{(x-1)(x+1)} = 2^{x^2 - 1} \). 3. **Now rewrite the expression**: - The full expression becomes: \[ \frac{2^{x+}}{2^{x^2 - x}} \div \frac{2^{2x + 2}}{2^{x^2 - 1}}. \] 4. **Convert the division into multiplication**: - Recall that dividing by a fraction is equivalent to multiplying by its reciprocal: \[ = \frac{2^{x+}}{2^{x^2 - x}} \cdot \frac{2^{x^2 - 1}}{2^{2x + 2}}. \] 5. **Combine the fractions**: - Combine the first part: \[ = 2^{x+ - (x^2 - x)} \cdot 2^{(x^2 - 1) - (2x + 2)}. \] 6. **Simplify the exponents**: - The exponent from the first part becomes: \[ x+ - (x^2 - x) = x + -x^2 + x = -x^2 + 2x. \] - The exponent from the second part becomes: \[ (x^2 - 1) - (2x + 2) = x^2 - 1 - 2x - 2 = x^2 - 2x - 3. \] 7. **Final combination**: - Now, we combine both parts: \[ 2^{-x^2 + 2x} \cdot 2^{x^2 - 2x - 3} = 2^{(-x^2 + 2x) + (x^2 - 2x - 3)} = 2^{-3}. \] 8. **Conclusion**: - Thus, the simplified expression is: \[ \frac{1}{8}. \] Therefore, the final answer is \( \frac{1}{8} \).