اختبار اختيار من متعدد تفاعلي: Polynomial/Rational Inequalities Review - Reveal

To solve \(x^2 - x \leq 6\), first rewrite it as \(x^2 - x - 6 \leq 0\). Factoring the quadratic gives \((x - 3)(x + 2) \leq 0\). The critical points are x = 3 and x = -2. Testing the intervals shows the expression is negative or zero between these roots. Since the inequality is \(\leq\), we include the endpoints, resulting in [-2, 3].

Factoring the quadratic 2x² + 9x - 5 < 0 results in (2x - 1)(x + 5) < 0. The roots are \(x = \frac{1}{2}\) and x = -5. The expression is negative in the interval between these two roots. Because it is a strict inequality (<), we do not include the endpoints, giving the interval \((-5, \frac{1}{2})\).

The critical points are x = -6 (from the numerator) and x = -3 (from the denominator). Testing values in the intervals shows the expression is positive between -6 and -3. Since the inequality is strict (>), we use parentheses, and since the denominator cannot be zero, x = -3 is excluded. The solution is (-6, -3).

First, simplify the expression: \(\frac{3x - (x + 4)}{x + 4} \geq 0 \Rightarrow \frac{2x - 4}{x + 4} \geq 0 \Rightarrow \frac{2(x - 2)}{x + 4} \geq 0\). The critical points are x = 2 and x = -4. The fraction is positive or zero when x < -4 or \(x \geq 2\). Note that x = -4 must be excluded because it is in the denominator, while x = 2 is included. Thus, \((-\infty, -4) \cup [2, \infty)\).

Subtract 1 from both sides: \(\frac{3x - 7}{x + 2} - 1 > 0 \Rightarrow \frac{3x - 7 - (x + 2)}{x + 2} > 0 \Rightarrow \frac{2x - 9}{x + 2} > 0\). The critical points are x = 4.5 (or \(\frac{9}{2}\)) and x = -2. Testing the sign of the fraction shows it is positive for values less than -2 or greater than 4.5. Since it is a strict inequality (>), endpoints are excluded, resulting in \((-\infty, -2) \cup (\frac{9}{2}, \infty)\).