9. \( \frac{x^{2}-4}{x} \div \frac{2 x^{2}+8}{x} \div \frac{4 x-2 x^{2}}{x} \)

To solve \( \frac{x^{2}-4}{x} \div \frac{2 x^{2}+8}{x} \div \frac{4 x-2 x^{2}}{x} \), we'll start by simplifying each fraction separately. 1. The first fraction, \( \frac{x^{2}-4}{x} \), can be factored as: \[ \frac{(x-2)(x+2)}{x} \] 2. The second fraction, \( \frac{2x^{2}+8}{x} \), can be simplified by factoring out the common term: \[ \frac{2(x^{2}+4)}{x} = \frac{2(x^{2}+4)}{x} \] 3. The third fraction, \( \frac{4x-2x^{2}}{x} \), can also be simplified: \[ \frac{2x(2-x)}{x} = 2(2 - x) \] Now we rewrite the expression with these simplifications: \[ \frac{(x-2)(x+2)}{x} \div \frac{2(x^{2}+4)}{x} \div (2(2-x)) \] Now, division by a fraction is equivalent to multiplication by its reciprocal: So we have: \[ \frac{(x - 2)(x + 2)}{x} \times \frac{x}{2(x^2 + 4)} \times \frac{1}{2(2 - x)} \] Now simplifying this gives: \[ = \frac{(x - 2)(x + 2)}{2(x^2 + 4)(2 - x)} \] Notice that \( (2 - x) = -1(x - 2) \), thus: \[ = \frac{(x - 2)(x + 2)}{2(x^2 + 4)(-1)(x - 2)} \] If \( x \neq 2 \), we can cancel \( (x - 2) \): \[ = \frac{-(x + 2)}{2(x^2 + 4)} \] Thus, the simplified result of the expression is: \[ -\frac{x + 2}{2(x^2 + 4)} \]