Ejercicio 164 – Solucionario Baldor

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DESCARGA

CAPITULO XIX

Inecuaciones

Ejercicio 164

Hallar el límite de x en las inecuaciones siguientes:

$\begin{matrix} x - 5 & < 2 x - 6 \\ x - 2 x & < - 6 + 5 \\ - x & < - 1 \\ x & > 1 \end{matrix}$
$\begin{matrix} 5 x - 12 & > 3 x - 4 \\ 5 x - 3 x & > 12 - 4 \\ 2 x & > 8 \\ x & > \frac{}{2} \\ x & > 4 \end{matrix}$
$\begin{matrix} x - 6 & > 21 - 8 x \\ x + 8 x & > 21 + 6 \\ 9 x & > 27 \\ x & > \frac{}{9} \\ x & > 3 \end{matrix}$
$\begin{matrix} 3 x - 14 & < 7 x - 2 \\ 3 x - 7 x & < 14 - 2 \\ - 4 x & < 12 \\ x & > - \frac{}{4} \\ x & > - 3 \end{matrix}$
$\begin{matrix} 2 x - \frac{5}{3} & > \frac{x}{3} + 10 \\ 2 x - \frac{x}{3} & > 10 + \frac{5}{3} \\ \frac{6 x - x}{3} & > \frac{30 + 5}{3} \\ 5 x & > 35 \\ x & > \frac{}{5} \\ x & > 7 \end{matrix}$
$\begin{matrix} 3 x - 4 + \frac{x}{4} & < \frac{5 x}{2} + 2 \\ 3 x + \frac{x}{4} - \frac{5 x}{2} & < 4 + 2 \\ \frac{12 x + x - 10 x}{4} & < 6 \\ \frac{3 x}{4} & < 6 \\ x & < \frac{4}{3} \\ x & < 8 \end{matrix}$
$\begin{matrix} {(x - 1)}^{2} - 7 & > {(x - 2)}^{2} \\ {(x - 1)}^{2} - {(x - 2)}^{2} & > 7 \\ [(x - 1) + (x - 2)] [(x - 1) - (x - 2)] & > 7 \\ [x - 1 + x - 2] [x - 1 - x + 2] & > 7 \\ [2 x - 3] [x - 1 - x + 2] & > 7 \\ 2 x - 3 & > 7 \\ 2 x & > 7 + 3 \\ x & > \frac{}{2} \\ x & > 5 \end{matrix}$
$\begin{matrix} (x + 2) (x - 1) + 26 & < (x + 4) (x + 5) \\ x^{2} + x - 2 + 26 & < x^{2} + 9 x + 20 \\ x - 9 x & < 20 - 24 \\ - 8 x & < - 4 \\ 8 x & > 4 \\ x & > \frac{4}{} \\ x & > \frac{1}{2} \end{matrix}$
$\begin{matrix} 3 (x - 2) + 2 x (x + 3) & > (2 x - 1) (x + 4) \\ 3 x - 6 + 2 x^{2} + 6 x & > 2 x^{2} + 8 x - x - 4 \\ 9 x - 6 & > 7 x - 4 \\ 9 x - 7 x & > 6 - 4 \\ 2 x & > 2 \\ x & > \frac{2}{2} \\ x & > 1 \end{matrix}$
$\begin{matrix} 6 (x^{2} + 1) - (2 x - 4) (3 x + 2) & < 3 (5 x + 21) \\ 6 x^{2} + 6 - (6 x^{2} + 4 x - 12 x - 8) & < 15 x + 63 \\ 6 x^{2} + 6 - 6 x^{2} - 4 x + 12 x + 8 & < 15 x + 63 \\ 8 x + 14 & < 15 x + 63 \\ 8 x - 15 x & < 63 - 14 \\ - 7 x & < 49 \\ x & > - \frac{}{7} \\ x & > - 7 \end{matrix}$
$\begin{matrix} (x - 4) (x + 5) & < (x - 3) (x - 2) \\ x^{2} + x - 20 & < x^{2} - 5 x + 6 \\ x + 5 x & < 20 + 6 \\ 6 x & < 26 \\ x & < \frac{}{} \\ x & < \frac{13}{3} \end{matrix}$
$\begin{matrix} {(2 x - 3)}^{2} + 4 x^{2} (x - 7) & < 4 {(x - 2)}^{3} \\ 4 x^{2} - 12 x + 9 + 4 x^{3} - 28 x^{2} & < 4 (x^{2} - 4 x + 4) (x - 2) \\ 4 x^{3} - 24 x^{2} - 12 x + 9 & < 4 (x^{3} - 4 x^{2} + 4 x - 2 x^{2} + 8 x - 8) \\ 4 x^{3} - 24 x^{2} - 12 x + 9 & < 4 (x^{3} - 6 x^{2} + 12 x - 8) \\ 4 x^{3} - 24 x^{2} - 12 x + 9 & < 4 x^{3} - 24 x^{2} + 48 x - 32 \\ - 12 x - 48 x & < - 9 - 32 \\ - 60 x & < - 41 \\ 60 x & > 41 \\ x & > \frac{41}{60} \end{matrix}$
$\begin{matrix} \frac{2 x + 1}{3 x - 1} & > \frac{2 x + 5}{3 x + 2} \\ (2 x + 1) (3 x + 2) & > (2 x + 5) (3 x - 1) \\ 6 x^{2} + 4 x + 3 x + 2 & > 6 x^{2} - 2 x + 15 x - 5 \\ 7 x + 2 & > 13 x - 5 \\ 7 x - 13 x & > - 2 - 5 \\ - 6 x & > - 7 \\ 6 x & < 7 \\ x & < \frac{7}{6} \end{matrix}$
$\begin{matrix} \frac{x + 3}{3} - \frac{4}{x + 2} & > \frac{x}{3} \\ \frac{(x + 2) (x + 3) - 12}{3 (x + 2)} & > \frac{x}{3} \\ (x + 2) (x + 3) - 12 & > x (x + 2) \\ x^{2} + 5 x + 6 - 12 & > x^{2} + 2 x \\ 5 x - 2 x & > 6 \\ 3 x & > 6 \\ x & > \frac{}{3} \\ x & > 2 \end{matrix}$
$\begin{matrix} \frac{5}{3 x + 1} - \frac{20}{9 x^{2} - 1} & < \frac{2}{3 x - 1} \\ \frac{5}{3 x + 1} - \frac{20}{(3 x + 1) (3 x - 1)} & < \frac{2}{3 x - 1} \\ \frac{5 (3 x - 1) - 20}{(3 x + 1) (3 x - 1)} & < \frac{2}{3 x - 1} \\ 5 (3 x - 1) - 20 & < 2 (3 x + 1) \\ 15 x - 5 - 20 & < 6 x + 2 \\ 15 x - 25 & < 6 x + 2 \\ 15 x - 6 x & < 25 + 2 \\ 9 x & < 27 \\ x & < \frac{}{9} \\ x & < 3 \end{matrix}$
$\begin{matrix} \frac{1}{x^{2} + x} & > \frac{1}{x^{2} - x} - \frac{1}{x^{2} - 1} \\ \frac{1}{x (x + 1)} & > \frac{1}{x (x - 1)} - \frac{1}{(x - 1) (x + 1)} \\ \frac{1}{x (x + 1)} & > \frac{x + 1 - x}{x (x + 1) (x - 1)} \\ 1 & > \frac{1}{x - 1} \\ x - 1 & > 1 \\ x & > 1 + 1 \\ x & > 2 \end{matrix}$
Hallar los números cuyo tercio aumentado en 15 sea mayor que su mitad aumentada en 1
$\begin{matrix} x & n ú m e r o s e n t e r o s \\ \frac{x}{3} + 15 & > \frac{x}{2} + 1 \\ \frac{x}{3} - \frac{x}{2} & > - 15 + 1 \\ \frac{2 x - 3 x}{6} & > - 14 \\ - x & > 6 (- 14) \\ x & < 84 \\ N ú m e r o s e n t e r o s m e n o r e s q u e 84 \end{matrix}$