Wheeler Gardner
08/14/2023 · Junior High School

Resolver en \( \mathbb{R} \), indicando el dominio respectivo: [25 puntos] \( \begin{array}{lll}\text { (a) }-2 x^{2}-2 x<4\left(-3-\frac{x}{2}\right) & \text { (b) } \frac{3}{4 x-8} \geq \frac{x}{x-2} & \text { (c) } \frac{4 x}{x+\frac{9}{2}}=-\frac{x^{2}}{x+\frac{9}{2}}-2\end{array} \)

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Para resolver cada una de las desigualdades dadas, primero simplificaremos las expresiones y luego resolveremos las desigualdades indicando el dominio respectivo. (a) \( -2x^{2} - 2x < 4(-3 - \frac{x}{2}) \) Simplificando la expresión: \[ -2x^{2} - 2x < -12 - 2x \] Sumando \( 2x \) a ambos lados: \[ -2x^{2} < -12 \] Dividiendo por -2 (recordando que al dividir por un número negativo, el signo de la desigualdad cambia): \[ x^{2} > 6 \] Por lo tanto, la solución de la desigualdad es \( x \in (-\infty, -\sqrt{6}) \cup (\sqrt{6}, \infty) \). (b) \( \frac{3}{4x-8} \geq \frac{x}{x-2} \) Multiplicando por \( (4x-8)(x-2) \) para eliminar los denominadores: \[ 3(x-2) \geq x(4x-8) \] Expandiendo y simplificando: \[ 3x - 6 \geq 4x^{2} - 8x \] Moviendo todos los términos a un lado: \[ 4x^{2} - 11x + 6 \leq 0 \] Factorizando: \[ (4x-3)(x-2) \leq 0 \] La solución de la desigualdad es \( x \in \left[\frac{3}{4}, 2\right] \). (c) \( \frac{4x}{x+\frac{9}{2}} = -\frac{x^{2}}{x+\frac{9}{2}} - 2 \) Multiplicando por \( x+\frac{9}{2} \) para eliminar los denominadores: \[ 4x = -x^{2} - 2(x+\frac{9}{2}) \] Expandiendo y simplificando: \[ 4x = -x^{2} - 2x - 9 \] Moviendo todos los términos a un lado: \[ x^{2} + 6x + 9 = 0 \] Factorizando: \[ (x+3)^{2} = 0 \] Por lo tanto, la solución de la desigualdad es \( x = -3 \). En resumen: (a) \( x \in (-\infty, -\sqrt{6}) \cup (\sqrt{6}, \infty) \) (b) \( x \in \left[\frac{3}{4}, 2\right] \) (c) \( x = -3 \)

Quick Answer

(a) \( x \in (-\infty, -\sqrt{6}) \cup (\sqrt{6}, \infty) \) (b) \( x \in \left[\frac{3}{4}, 2\right] \) (c) \( x = -3 \)
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