1)5х-3>4 2)4(х-4)≤12 3)(х+2)^2(х-3)/х-1 ≤0 4)|2х-4|<7 5)4х+2<7 6)3(х+2)≥11 7)(х-4)^2(х+3)/(х-2) ≥0 8)|3х+1|>4

Let's solve each inequality step by step. ### 1) \( 5x - 3 > 4 \) 1. Add 3 to both sides: \[ 5x > 7 \] 2. Divide by 5: \[ x > \frac{7}{5} \] ### Solution: \( x > 1.4 \) --- ### 2) \( 4(x - 4) \leq 12 \) 1. Divide both sides by 4: \[ x - 4 \leq 3 \] 2. Add 4 to both sides: \[ x \leq 7 \] ### Solution: \( x \leq 7 \) --- ### 3) \( \frac{(x + 2)^2(x - 3)}{x - 1} \leq 0 \) First, find the critical points by setting the numerator and denominator to zero. - Numerator: \( (x + 2)^2(x - 3) = 0 \) → \( x = -2 \) (with multiplicity 2), \( x = 3 \) - Denominator: \( x - 1 = 0 \) → \( x = 1 \) Now, we test the intervals: (-∞, -2), (-2, 1), (1, 3), (3, ∞) 1. Choose a test point from each interval: - For \( x = -3 \): \( \frac{(x + 2)^2(x - 3)}{x - 1} = \frac{(1)^2(-6)}{-4} > 0 \) (interval is positive) - For \( x = 0 \): \( \frac{(2)^2(-3)}{-1} > 0 \) (interval is positive) - For \( x = 2 \): \( \frac{(4)^2(-1)}{1} < 0 \) (interval is negative) - For \( x = 4 \): \( \frac{(6)^2(1)}{3} > 0 \) (interval is positive) 2. The sign changes at the critical points and we consider the signs: - \( x = -2 \) is part of the solution since it's a squared term. - \( x = 1 \) is not included as it makes the expression undefined. - \( x = 3 \) is included. ### Solution: \( x \in [-2, 1) \cup [3, \infty) \) --- ### 4) \( |2x - 4| < 7 \) This means: 1. \( -7 < 2x - 4 < 7 \) Now solve each part. 1. \( 2x - 4 < 7 \): \[ 2x < 11 \implies x < \frac{11}{2} \] 2. \( -7 < 2x - 4 \): \[ -3 < 2x \implies x > -\frac{3}{2} \] ### Solution: \( -\frac{3}{2} < x < \frac{11}{2} \) --- ### 5) \( 4x + 2 < 7 \) 1. Subtract 2 from both sides: \[ 4x < 5 \] 2. Divide by 4: \[ x < \frac{5}{4} \] ### Solution: \( x < 1.25 \) --- ### 6) \( 3(x + 2) \geq 11 \) 1. Divide by 3: \[ x + 2 \geq \frac{11}{3} \] 2. Subtract 2: \[ x \geq \frac{11}{3} - 2 \implies x \geq \frac{5}{3} \] ### Solution: \( x \geq \frac{5}{3} \) --- ### 7) \( \frac{(x - 4)^2(x + 3)}{x - 2} \geq 0 \) First, find the critical points: - Numerator: \( (x - 4)^2(x + 3) = 0 \) → \( x = 4 \) (multiplicity 2), \( x = -3 \) - Denominator: \( x - 2 = 0 \) → \( x = 2 \) Testing intervals (-∞, -3), (-3, 2), (2, 4), (4, ∞): 1. For \( x = -4 \): positive 2. For \( x = 0 \): negative 3. For \( x = 3 \): negative 4. For \( x = 5 \): positive Since \( (x - 4)^2 \) is non-negative: - Included critical points -3 and 4. - Exclude 2 since it makes the expression undefined. ### Solution: \( x \in (-\infty, -3] \cup (2, 4] \) --- ### 8) \( |3x + 1| > 4 \) This means: 1. \( 3x + 1 < -4 \quad \text{or} \quad 3x + 1 > 4 \) **Solving the first case**: 1. \( 3x < -5 \implies x < -\frac{5}{3} \) **Solving the second case**: 1. \( 3x > 3 \implies x > 1 \) ### Solution: \( x < -\frac{5}{3} \quad \text{or} \quad x > 1 \) --- This concludes the solutions to all inequalities.