(vi) \( \frac{(16)^{x+1}+20\left(4^{2 x}\right)}{2^{x-3} \times 8^{x+2}} \)

Let’s simplify the expression \( \frac{(16)^{x+1}+20\left(4^{2x}\right)}{2^{x-3} \times 8^{x+2}} \). First, we can rewrite all the powers in terms of base 2: - \( 16 = 2^4 \), so \( (16)^{x+1} = (2^4)^{x+1} = 2^{4(x+1)} = 2^{4x + 4} \). - \( 4 = 2^2 \), so \( 4^{2x} = (2^2)^{2x} = 2^{4x} \), thus \( 20 \cdot 4^{2x} = 20 \cdot 2^{4x} = 20 \cdot 2^{4x} \). - We know \( 8 = 2^3 \), thus \( 8^{x+2} = (2^3)^{x+2} = 2^{3(x+2)} = 2^{3x + 6} \). Now substituting everything back into the expression, we get: \[ \frac{2^{4x + 4} + 20 \cdot 2^{4x}}{2^{x - 3} \cdot 2^{3x + 6}}. \] Combining the denominator, we find: \[ 2^{x-3 + 3x + 6} = 2^{4x + 3}. \] Now the expression simplifies to: \[ \frac{2^{4x + 4} + 20 \cdot 2^{4x}}{2^{4x + 3}}. \] Factoring \( 2^{4x} \) out of the numerator, we get: \[ \frac{2^{4x} (2^4 + 20)}{2^{4x + 3}} = \frac{2^{4x} (16 + 20)}{2^{4x + 3}} = \frac{2^{4x} \cdot 36}{2^{4x + 3}}. \] Now, we can simplify further: \[ = \frac{36}{2^3} = \frac{36}{8} = 4.5. \] Thus, the value of the original expression simplifies to: \[ \boxed{4.5}. \]