\( \frac{\text { Simplify }}{\frac{2 x}{2-x} \times \frac{6-3 x}{6} \div \frac{x+2}{4-x^{2}}} \)

To tackle the expression \( \frac{\text { Simplify }}{\frac{2 x}{2-x} \times \frac{6-3 x}{6} \div \frac{x+2}{4-x^{2}}} \), let's simplify it step by step. First, rewrite the division as multiplication by the reciprocal: \[ \frac{2 x}{2-x} \times \frac{6-3 x}{6} \times \frac{4-x^{2}}{x+2} \] Now, we can start simplifying each part: 1. **Simplify \( \frac{2 x}{2-x} \):** This stays as it is for now. 2. **Simplify \( \frac{6-3x}{6} \):** Factor out a 3 from the numerator: \[ \frac{6-3x}{6} = \frac{3(2-x)}{6} = \frac{2-x}{2} \] 3. **For \( \frac{4-x^{2}}{x+2} \)**: Recognize that \( 4 - x^2 \) is a difference of squares: \[ 4 - x^2 = (2-x)(2+x) \] So, this term becomes: \[ \frac{(2-x)(2+x)}{x+2} = (2-x) \cdot \frac{(2+x)}{(x+2)} \] Plugging everything back into the expression gives us: \[ \frac{2 x}{2-x} \times \frac{(2-x)}{2} \times (2-x) \cdot \frac{(2+x)}{(x+2)} \] Notice that the \( 2-x \) terms will cancel out: \[ = \frac{2 x}{2} \cdot (2+x) = x(2+x) = 2x+x^2 \] Thus, the final simplified form of the expression is: \[ \frac{x^2 + 2x}{1} \] So in conclusion, the simplified expression is: \[ x^2 + 2x \]