Simplify the following complex fractions: \( \begin{array}{ll}\text { (1) } \frac{2+\frac{1}{x}}{\frac{1}{x^{2}}} & \text { (2) } 3-\frac{x-\frac{9}{x}}{x-3}\end{array} \)

To tackle the complex fractions given, we'll simplify each one step by step. 1. For the first complex fraction \( \frac{2+\frac{1}{x}}{\frac{1}{x^{2}}} \): First, let's simplify the numerator \( 2 + \frac{1}{x} \). We can combine this into a single fraction: \[ 2 + \frac{1}{x} = \frac{2x}{x} + \frac{1}{x} = \frac{2x + 1}{x} \] Now, we rewrite the complex fraction: \[ \frac{\frac{2x + 1}{x}}{\frac{1}{x^2}} \] To divide by a fraction, we multiply by its reciprocal: \[ = \frac{2x + 1}{x} \cdot \frac{x^2}{1} = (2x + 1) \cdot x = 2x^2 + x \] So the simplified form is \( 2x^2 + x \). 2. For the second complex fraction \( 3 - \frac{x - \frac{9}{x}}{x - 3} \): First, we simplify the numerator \( x - \frac{9}{x} \): \[ x - \frac{9}{x} = \frac{x^2}{x} - \frac{9}{x} = \frac{x^2 - 9}{x} \] Rewrite the complex fraction: \[ 3 - \frac{\frac{x^2 - 9}{x}}{x - 3} \] Again, multiply by the reciprocal: \[ = 3 - \frac{x^2 - 9}{x(x - 3)} \] Next, \( x^2 - 9 \) can be factored as \( (x - 3)(x + 3) \): \[ = 3 - \frac{(x - 3)(x + 3)}{x(x - 3)} \] The \( (x - 3) \) terms cancel out (where \( x \neq 3 \)): \[ = 3 - \frac{x + 3}{x} \] Now, we rewrite \( 3 \): \[ 3 = \frac{3x}{x} \] Thus, we have: \[ = \frac{3x}{x} - \frac{x + 3}{x} = \frac{3x - (x + 3)}{x} = \frac{3x - x - 3}{x} = \frac{2x - 3}{x} \] So the simplified forms are: 1. \( 2x^2 + x \) 2. \( \frac{2x - 3}{x} \)