1. **State the problem:** Simplify the expressions: a) $\frac{6^n + 3^n}{2^{n+1} + 2}$ b) $\frac{12^x + 1}{6^{2x} + 3^x}$ 2. **Recall useful rules:** - Exponent rules: $a^{m+n} = a^m \cdot a^n$, $\left(a^m\right)^n = a^{mn}$ - Factorization: look for common bases or terms. 3. **Simplify (a):** Numerator: $6^n + 3^n$ Denominator: $2^{n+1} + 2 = 2 \cdot 2^n + 2 = 2(2^n + 1)$ Rewrite numerator: $6^n = (2 \cdot 3)^n = 2^n \cdot 3^n$ So numerator = $2^n 3^n + 3^n = 3^n (2^n + 1)$ Expression becomes: $$\frac{3^n (2^n + 1)}{2 (2^n + 1)}$$ Cancel common factor $2^n + 1$: $$\frac{3^n \cancel{(2^n + 1)}}{2 \cancel{(2^n + 1)}} = \frac{3^n}{2}$$ 4. **Simplify (b):** Numerator: $12^x + 1$ Denominator: $6^{2x} + 3^x$ Rewrite terms: $12^x = (6 \cdot 2)^x = 6^x \cdot 2^x$ $6^{2x} = (6^x)^2$ So denominator = $(6^x)^2 + 3^x$ Also, $3^x = (6^{x})^{\log_6 3}$ but better to express $3^x$ as is. Notice $3^x = (3)^x$ and $6^x = (2 \cdot 3)^x = 2^x 3^x$ Let's rewrite denominator: $6^{2x} + 3^x = (6^x)^2 + 3^x$ Substitute $6^x = 2^x 3^x$: Denominator = $(2^x 3^x)^2 + 3^x = 2^{2x} 3^{2x} + 3^x$ Factor $3^x$ from denominator: $3^x (2^{2x} 3^x + 1) = 3^x (2^{2x} 3^x + 1)$ Numerator is $12^x + 1 = 6^x 2^x + 1 = 2^x 3^x 2^x + 1 = 2^x 3^x 2^x + 1$ (rewriting is complex, better to keep as $12^x + 1$) Alternatively, rewrite numerator as $12^x + 1 = (12)^x + 1$ Let's try to factor denominator differently: Denominator: $6^{2x} + 3^x = (6^x)^2 + 3^x$ Write $3^x = (6^x)^{\log_6 3}$, but this is complicated. Instead, try to express numerator and denominator in terms of $3^x$ and $2^x$: Numerator: $12^x + 1 = (2^2 \cdot 3)^x + 1 = 2^{2x} 3^x + 1$ Denominator: $6^{2x} + 3^x = (2 \cdot 3)^{2x} + 3^x = 2^{2x} 3^{2x} + 3^x$ Factor $3^x$ from denominator: $3^x (2^{2x} 3^x + 1)$ So expression is: $$\frac{2^{2x} 3^x + 1}{3^x (2^{2x} 3^x + 1)}$$ Cancel common factor $2^{2x} 3^x + 1$: $$\frac{\cancel{2^{2x} 3^x + 1}}{3^x \cancel{2^{2x} 3^x + 1}} = \frac{1}{3^x}$$ 5. **Final answers:** a) $\boxed{\frac{3^n}{2}}$ b) $\boxed{\frac{1}{3^x}}$