Long Byrd
07/15/2024 · Junior High School

encontradá Ejercicios. A. \( \left|\frac{x}{3}-\frac{2}{3}\right|<2 \)

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Para resolver la desigualdad \( \left|\frac{x}{3}-\frac{2}{3}\right|<2 \), podemos seguir estos pasos: 1. **Eliminar el valor absoluto**: La desigualdad \( |A| < B \) se puede reescribir como \( -B < A < B \). En este caso, \( A = \frac{x}{3} - \frac{2}{3} \) y \( B = 2 \). Entonces, tenemos: \[ -2 < \frac{x}{3} - \frac{2}{3} < 2 \] 2. **Resolver la desigualdad**: Ahora resolveremos ambas partes de la desigualdad. - Para la parte izquierda: \[ -2 < \frac{x}{3} - \frac{2}{3} \] Sumamos \( \frac{2}{3} \) a ambos lados: \[ -2 + \frac{2}{3} < \frac{x}{3} \] Para sumar \( -2 \) y \( \frac{2}{3} \), convertimos \( -2 \) a fracción: \[ -2 = -\frac{6}{3} \] Entonces: \[ -\frac{6}{3} + \frac{2}{3} = -\frac{4}{3} \] Así que tenemos: \[ -\frac{4}{3} < \frac{x}{3} \] Multiplicamos ambos lados por 3: \[ -4 < x \] - Para la parte derecha: \[ \frac{x}{3} - \frac{2}{3} < 2 \] Sumamos \( \frac{2}{3} \) a ambos lados: \[ \frac{x}{3} < 2 + \frac{2}{3} \] Convertimos \( 2 \) a fracción: \[ 2 = \frac{6}{3} \] Entonces: \[ 2 + \frac{2}{3} = \frac{6}{3} + \frac{2}{3} = \frac{8}{3} \] Así que tenemos: \[ \frac{x}{3} < \frac{8}{3} \] Multiplicamos ambos lados por 3: \[ x < 8 \] 3. **Combinar las soluciones**: Ahora combinamos las dos partes de la desigualdad: \[ -4 < x < 8 \] Por lo tanto, la solución de la desigualdad \( \left|\frac{x}{3}-\frac{2}{3}\right|<2 \) es: \[ x \in (-4, 8) \]

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La solución de la desigualdad \( \left|\frac{x}{3}-\frac{2}{3}\right|<2 \) es \( x \in (-4, 8) \).
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