5. \( \frac{x^{2}-x}{x^{2}-2 x-3} \div \frac{x^{2}-4 x+3}{1-x^{2}} \)
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To simplify \( \frac{x^{2}-x}{x^{2}-2 x-3} \div \frac{x^{2}-4 x+3}{1-x^{2}} \), we first rewrite the division as multiplication by the reciprocal: \[ \frac{x^{2}-x}{x^{2}-2 x-3} \cdot \frac{1-x^{2}}{x^{2}-4 x+3} \] Next, we'll factor each polynomial: 1. **Numerator of the first fraction**: \( x^{2}-x = x(x-1) \) 2. **Denominator of the first fraction**: \( x^{2}-2x-3 = (x-3)(x+1) \) 3. **Numerator of the second fraction (reciprocal)**: \( 1-x^{2} = -(x^{2}-1) = -(x-1)(x+1) \) 4. **Denominator of the second fraction**: \( x^{2}-4x+3 = (x-3)(x-1) \) Now substituting the factored forms into our expression: \[ \frac{x(x-1)}{(x-3)(x+1)} \cdot \frac{-(x-1)(x+1)}{(x-3)(x-1)} \] Now, we can simplify: Cancel \( (x-1) \) in the numerator and denominator: \[ = \frac{x \cdot -1 \cdot (x+1)}{(x-3)(x+1)(x-3)} \] Now we can cancel \( (x+1) \): \[ = \frac{-x}{(x-3)(x-3)} \] Thus, we finalize our simplification to: \[ \frac{-x}{(x-3)^{2}}, \quad x \neq 3, -1, \text{and } 1 \] Now, the final result is: \[ \frac{-x}{(x-3)^2} \] where \( x \) cannot be \( 3 \), \( 1 \), or \(-1\).
