7) \( \frac{x^{2}-x-2}{x+1}+1=0 \)
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To solve the equation \( \frac{x^{2}-x-2}{x+1}+1=0 \), first, we can simplify the left-hand side. Start by rewriting the equation: \[ \frac{x^{2}-x-2}{x+1} = -1 \] Next, to eliminate the fraction, multiply both sides by \( x+1 \) (noting that \( x \neq -1 \)): \[ x^{2}-x-2 = -1(x+1) \] This simplifies to: \[ x^{2}-x-2 = -x - 1 \] Now, combine like terms: \[ x^{2}-x + x - 2 + 1 = 0 \implies x^{2} - 1 = 0 \] Now we can factor the quadratic: \[ (x-1)(x+1) = 0 \] Setting each factor equal to zero gives us the solutions: \[ x - 1 = 0 \quad \text{or} \quad x + 1 = 0 \] \[ x = 1 \quad \text{or} \quad x = -1 \] However, since we multiplied by \( x + 1 \), we must exclude \( x = -1 \) since it would make the original denominator zero. Thus, the only solution is: \[ \boxed{1} \]