1) \( \frac{4^{x}-2^{x+2}+3}{9-4^{x}} \)

Let's rewrite \( 4^x \) in terms of \( 2^x \). Since \( 4^x = (2^2)^x = (2^x)^2 \), we can set \( y = 2^x \). This transforms the expression into: \[ \frac{y^2 - 2^{x+2} + 3}{9 - y^2} \] Now, substituting \( 2^{x+2} = 4y \), we get: \[ \frac{y^2 - 4y + 3}{9 - y^2} \] Factoring the numerator: \[ y^2 - 4y + 3 = (y - 1)(y - 3) \] So our expression now looks like this: \[ \frac{(y - 1)(y - 3)}{9 - y^2} \] The denominator can be rewritten as: \[ 9 - y^2 = (3 - y)(3 + y) \] Therefore, our expression simplifies to: \[ \frac{(y - 1)(y - 3)}{(3 - y)(3 + y)} \] Now it's important to notice that \( 3 - y = -(y - 3) \): \[ = \frac{(y - 1)(y - 3)}{-(y - 3)(3 + y)} = -\frac{(y - 1)}{(3 + y)} \quad \text{for } y \neq 3 \] So, the final simplified expression is: \[ -\frac{y - 1}{3 + y} \] Finally, substituting back \( y = 2^x \): \[ -\frac{2^x - 1}{3 + 2^x} \]