$\dfrac{9 (4^x)^2}{16^{x + 1} - 2^{x + 1} \times 8^x} =$

1. 0

2. 1

3. $\dfrac{14}{9}$

4. $\dfrac{9}{14}$

Given,

$\dfrac{9 (4^x)^2}{16^{x + 1} - 2^{x + 1} \times 8^x}$

Solving Numerator:

⇒ 9 × (4^x)²

⇒ 9 × [(2²)^x]²

⇒ 9 × 2^4x

Solving Denominator:

⇒ 16^{x + 1} - 2^{x + 1} × 8^x

⇒ (2⁴)^{x + 1} - 2^{x + 1} × (2³)^x

⇒ 2^{4x + 4} - 2^{x + 1} × 2^3x

⇒ 2^{4x + 4} - 2^{x + 1 + 3x}

⇒ 2^{4x + 4} - 2^{4x + 1}

Substituting the simplified numerator and denominator in the original expression:

$\Rightarrow \dfrac{9 \times 2^{4x}}{2^{4x + 4} - 2^{4x + 1}} \\[1em] \Rightarrow \dfrac{9 \times 2^{4x}}{2^{4x}.2^4 - 2^{4x}.2^1} \\[1em] \Rightarrow \dfrac{9 \times 2^{4x}}{2^{4x}(2^4 - 2^1)} \\[1em] \Rightarrow \dfrac{9 \times 2^{4x}}{2^{4x} \times (16 - 2)} \\[1em] \Rightarrow \dfrac{9}{14}.$

Hence, option 4 is the correct option.