26. \( \left[\frac{|x-2|+3}{2} \rrbracket \geqslant 5\right. \) 27. \( \llbracket \frac{5+x}{5-x} \rrbracket \leqslant 1 \) 28. \( \llbracket 4 x^{2}-5 x-4 \rrbracket \leqslant 1 \)

To solve the inequalities given, we will analyze each one step by step. ### Problem 26: \[ \left[\frac{|x-2|+3}{2} \rrbracket \geqslant 5 \] First, we can rewrite the inequality: \[ \frac{|x-2| + 3}{2} \geq 5 \] Multiplying both sides by 2: \[ |x-2| + 3 \geq 10 \] Subtracting 3 from both sides: \[ |x-2| \geq 7 \] This absolute value inequality can be split into two cases: 1. \( x - 2 \geq 7 \) which simplifies to \( x \geq 9 \) 2. \( x - 2 \leq -7 \) which simplifies to \( x \leq -5 \) Thus, the solution set for this inequality is: \[ x \leq -5 \quad \text{or} \quad x \geq 9 \] ### Problem 27: \[ \llbracket \frac{5+x}{5-x} \rrbracket \leqslant 1 \] This means we need to find when: \[ \frac{5+x}{5-x} \leq 1 \] To solve this, we can first rearrange the inequality: \[ 5+x \leq 5-x \] Adding \( x \) to both sides: \[ 5 + 2x \leq 5 \] Subtracting 5 from both sides: \[ 2x \leq 0 \] Dividing by 2: \[ x \leq 0 \] Next, we also need to consider when the expression is defined. The expression \( \frac{5+x}{5-x} \) is undefined when \( 5 - x = 0 \) or \( x = 5 \). Thus, we need to ensure \( x \neq 5 \). The solution set for this inequality is: \[ x \leq 0 \] ### Problem 28: \[ \llbracket 4x^{2}-5x-4 \rrbracket \leqslant 1 \] This means we need to solve: \[ 4x^2 - 5x - 4 \leq 1 \] Rearranging gives: \[ 4x^2 - 5x - 5 \leq 0 \] Next, we can find the roots of the quadratic equation \( 4x^2 - 5x - 5 = 0 \) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 4 \cdot (-5)}}{2 \cdot 4} \] \[ = \frac{5 \pm \sqrt{25 + 80}}{8} = \frac{5 \pm \sqrt{105}}{8} \] Let \( r_1 = \frac{5 - \sqrt{105}}{8} \) and \( r_2 = \frac{5 + \sqrt{105}}{8} \). The quadratic opens upwards (since the coefficient of \( x^2 \) is positive), so the solution to \( 4x^2 - 5x - 5 \leq 0 \) is: \[ r_1 \leq x \leq r_2 \] ### Summary of Solutions: 1. For Problem 26: \( x \leq -5 \) or \( x \geq 9 \) 2. For Problem 27: \( x \leq 0 \) 3. For Problem 28: \( \frac{5 - \sqrt{105}}{8} \leq x \leq \frac{5 + \sqrt{105}}{8} \)