$$ \begin{array}{ll} ext { 1.1.2. } & \frac{15^{n+2} \cdot 45^{1-n}}{3^{3-n}} \ ext { 1.1.3 } & \frac{2^{x}+2^{x+1}}{2^{x}-2^{x+2}}\end{array} $$

Question

$$ \begin{array}{ll}	ext { 1.1.2. } & \frac{15^{n+2} \cdot 45^{1-n}}{3^{3-n}} \ 	ext { 1.1.3 } & \frac{2^{x}+2^{x+1}}{2^{x}-2^{x+2}}\end{array} $$

UpStudy AI · Accepted Answer

$$
\textbf{1.1.2: } \frac{15^{n+2} \cdot 45^{1-n}}{3^{3-n}}
$$

Write $$15$$ and $$45$$ in terms of their prime factors:
$$
15 = 3 \cdot 5,\quad 45 = 3^2 \cdot 5.
$$

Rewrite the numerator:
$$
15^{n+2} = (3\cdot5)^{n+2} = 3^{n+2} \cdot 5^{n+2},
$$
$$
45^{1-n} = (3^2\cdot5)^{1-n} = 3^{2(1-n)} \cdot 5^{1-n}.
$$

Multiply these:
$$
15^{n+2} \cdot 45^{1-n} = 3^{n+2} \cdot 5^{n+2} \cdot 3^{2(1-n)} \cdot 5^{1-n}.
$$

Combine the powers of $$3$$ and $$5$$:
$$
3^{n+2} \cdot 3^{2(1-n)} = 3^{n+2 + 2(1-n)} = 3^{n+2 + 2 - 2n} = 3^{4-n},
$$
$$
5^{n+2} \cdot 5^{1-n} = 5^{n+2+1-n} = 5^{3}.
$$

Thus, the numerator becomes:
$$
3^{4-n} \cdot 5^{3}.
$$

Now, divide by the denominator:
$$
\frac{3^{4-n} \cdot 5^{3}}{3^{3-n}}.
$$

Simplify the powers of $$3$$:
$$
\frac{3^{4-n}}{3^{3-n}} = 3^{(4-n) - (3-n)} = 3^{1} = 3.
$$

So the entire expression becomes:
$$
3 \cdot 5^3 = 3 \cdot 125 = 375.
$$

$$
\textbf{1.1.3: } \frac{2^{x}+2^{x+1}}{2^{x}-2^{x+2}}
$$

Factor $$2^x$$ from each term. For the numerator:
$$
2^{x}+2^{x+1} = 2^x (1+2) = 2^x \cdot 3.
$$

For the denominator:
$$
2^{x}-2^{x+2} = 2^x (1-2^2) = 2^x (1-4) = 2^x \cdot (-3).
$$

Now, the fraction becomes:
$$
\frac{2^x \cdot 3}{2^x \cdot (-3)}.
$$

Cancel the common factor $$2^x$$ (assuming $$2^x \neq 0$$):
$$
\frac{3}{-3} = -1.
$$

Thus, the simplified results are:

1. $$ \frac{15^{n+2} \cdot 45^{1-n}}{3^{3-n}} = 375 $$
2. $$ \frac{2^{x}+2^{x+1}}{2^{x}-2^{x+2}} = -1 $$
1. $$ \frac{15^{n+2} \cdot 45^{1-n}}{3^{3-n}} = 375 $$ 2. $$ \frac{2^{x}+2^{x+1}}{2^{x}-2^{x+2}} = -1 $$

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