Evans Ingram
12/12/2023 · Senior High School

Resolver en \( \mathbb{R} \), indicando el dominio respectivo: \( \begin{array}{ll}\text { (a) }-2 x^{2}-2 x<4\left(-3-\frac{x}{2}\right) & \text { (b) } \frac{3}{4 x-8} \geq \frac{x}{x-2}\end{array} \)

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Claro, aquí tienes las soluciones para los problemas dados: ### (a) \( -2x^2 - 2x < 4\left(-3 - \frac{x}{2}\right) \) 1. **Expande y reordena:** \[ -2x^2 - 2x < -12 - 2x \] 2. **Simplifica:** \[ -2x^2 < -12 \] 3. **Divide por -2 (recuerda cambiar el signo):** \[ x^2 > 6 \] 4. **Raíz cuadrada:** \[ x > \sqrt{6} \quad \text{o} \quad x < -\sqrt{6} \] 5. **Dominio:** \[ (-\infty, -\sqrt{6}) \cup (\sqrt{6}, \infty) \] ### (b) \( \frac{3}{4x - 8} \geq \frac{x}{x - 2} \) 1. **Encontrar el dominio:** \[ 4x - 8 \neq 0 \quad \text{y} \quad x - 2 \neq 0 \] \[ x \neq 2 \quad \text{y} \quad x \neq 2 \] 2. **Multiplica por \( (4x - 8)(x - 2) \) (considerando el signo correcto):** \[ 3(x - 2) \geq x(4x - 8) \] 3. **Expande y reordena:** \[ 3x - 6 \geq 4x^2 - 8x \] \[ 4x^2 - 11x + 6 \leq 0 \] 4. **Factoriza:** \[ (4x - 3)(x - 2) \leq 0 \] 5. **Encontrar los puntos críticos:** \[ x = \frac{3}{4} \quad \text{y} \quad x = 2 \] 6. **Determinar el intervalo:** \[ \frac{3}{4} \leq x \leq 2 \] 7. **Dominio:** \[ \left[\frac{3}{4}, 2\right] \] Espero que esto te sea útil. ¡Buena suerte con tu estudio!

Quick Answer

### (a) \( -2x^2 - 2x < 4\left(-3 - \frac{x}{2}\right) \) Dominio: \( (-\infty, -\sqrt{6}) \cup (\sqrt{6}, \infty) \) ### (b) \( \frac{3}{4x - 8} \geq \frac{x}{x - 2} \) Dominio: \( \left[\frac{3}{4}, 2\right] \)
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