1.2 Simplify the following: $$ 1.2 .1 \frac{2^{a+1}-2^{a-1}}{2^{a}} $$ $$ 1.2 .2 \quad \frac{x^{2}-x+1}{x^{3}+1} \div \frac{2 x}{2 x+2} $$

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$$
\textbf{For 1.2.1:}
$$

Given the expression:
$$
\frac{2^{a+1}-2^{a-1}}{2^a}
$$

1. Rewrite the terms in the numerator:
   $$
   2^{a+1} = 2\cdot2^a \quad \text{and} \quad 2^{a-1} = \frac{2^a}{2}
   $$
2. Substitute these into the numerator:
   $$
   2\cdot2^a - \frac{2^a}{2}
   $$
3. Factor out $$2^a$$:
   $$
   2^a\left(2 - \frac{1}{2}\right)
   $$
4. Simplify the expression inside the parentheses:
   $$
   2 - \frac{1}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2}
   $$
5. The expression becomes:
   $$
   \frac{2^a\left(\frac{3}{2}\right)}{2^a}
   $$
6. Cancel $$2^a$$ from the numerator and denominator:
   $$
   \frac{3}{2}
   $$

$$
\textbf{For 1.2.2:}
$$

Given the expression:
$$
\frac{x^{2}-x+1}{x^{3}+1} \div \frac{2x}{2x+2}
$$

1. Replace the division with multiplication by the reciprocal:
   $$
   \frac{x^2 - x + 1}{x^3 + 1} \times \frac{2x+2}{2x}
   $$
2. Factor the denominator $$x^{3}+1$$ as a sum of cubes:
   $$
   x^3+1 = (x+1)(x^2-x+1)
   $$
3. Factor $$2x+2$$:
   $$
   2x+2 = 2(x+1)
   $$
4. Substitute these factorizations into the expression:
   $$
   \frac{x^2 - x + 1}{(x+1)(x^2 - x + 1)} \times \frac{2(x+1)}{2x}
   $$
5. Cancel the common factor $$x^2 - x + 1$$ in the numerator and denominator:
   $$
   \frac{1}{x+1} \times \frac{2(x+1)}{2x}
   $$
6. Cancel the common factor $$x+1$$:
   $$
   \frac{2}{2x}
   $$
7. Simplify the resulting fraction:
   $$
   \frac{2}{2x} = \frac{1}{x}
   $$

$$
\boxed{\frac{3}{2}, \quad \frac{1}{x}}
$$
$$ \frac{3}{2}, \quad \frac{1}{x} $$

1.2 Simplify the following: \( 1.2 .1 \frac{2^{a+1}-2^{a-1}}{2^{a}} \) \( 1.2 .2 \quad \frac{x^{2}-x+1}{x^{3}+1} \div \frac{2 x}{2 x+2} \)

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