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: \( x > 1.4 \)

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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: \( x \leq 7 \)

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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: \( x \in [-2, 1) \cup [3, \infty) \)

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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: \( -\frac{3}{2} < x < \frac{11}{2} \)

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5) \( 4x + 2 < 7 \)

1. Subtract 2 from both sides: \[ 4x < 5 \]
2. Divide by 4: \[ x < \frac{5}{4} \]

Solution: \( x < 1.25 \)

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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: \( x \geq \frac{5}{3} \)

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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: \( x \in (-\infty, -3] \cup (2, 4] \)

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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: \( x < -\frac{5}{3} \quad \text{or} \quad x > 1 \)

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This concludes the solutions to all inequalities.