Answer:
If \(x=-3\) then the denominator, \(x + 3 = 0\) and the fraction is undefined. ... Simplify: \[\frac{{x}^{2}-x-2}{{x}^{2}-4}\div\frac{{x}^{2}+x}{{x}^{2}+2x}, \quad \left(x\ne 0;x\ne ±2\right)\] ... \(\dfrac{3x+2}{x^2 - 6x + 8} \times \dfrac{x-2}{3x^2 + 8x + 4}\) ... We need to find the value of \(x\) that will make the denominator equal to \(\text{0}\).
Missing: [tex]
Step-by-step explanation:
