Valor absoluto
En los problemas del 1 al 9, resolver la ecuación dada.
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\[ \mid x - 5 \mid = 4 \]
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\[ \mid 2x + 1 \mid = x + 3 \]
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\[ \mid x - 2 \mid = 3x - 9 \]
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\[ \mid x-2 \mid = 9-3x \]
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\[ \mid x + 4 \mid = \mid 2 - x \mid \]
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\[ \mid x - 1 \mid = \mid 2x -4 \mid \]
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\[ \left| \frac{3x-2}{2} \right| = \mid x-4 \mid \]
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\[ \left| 5-\frac{2}{x} \right| = 3 \]
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\[ \left| \frac{x-5}{2x-3} \right| = 1 \]
En los problemas del 10 al 26, resolver la inecuación dada.
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\[ \mid x - 4 \mid < 3 \]
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\[ \mid 3x + 1 \mid < 15 \]
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\[ \left| \frac{2x}{3} -1 \right| < 2 \]
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\[ \mid -3x - 2 \mid \leq 4 \]
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\[ \mid 5x + 2 \mid \geq 1 \]
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\[ \mid -4x - 3 \mid > 1 \]
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\[ \left| \frac{2x}{5}-2 \right| \geq 3 \]
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\[ \mid x^2 - 5 \mid \geq 4 \]
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\[ 1 < \mid x \mid \leq 4 \]
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\[ 0 < \mid x - 3 \mid < 1 \]
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\[ \mid x - 1 \mid < \mid x \mid \]
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\[ \left| \frac{3-2x}{1+x} \right| \leq 1 \]
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\[ \left| \frac{1}{1-2x} \right| \geq \frac{1}{3} \]
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\[ \mid x - 1 \mid + \mid x - 2 \mid > 1 \]
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\[ \mid x - 1 \mid + \mid x + 1 \mid \leq 4 \]
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\[ \left| \frac{1}{2+x} \right| < \frac{1}{\mid x \mid} \]
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\[ \mid 3x - 5 \mid \leq \mid 2x - 1 \mid + \mid 2x + 3 \mid \]
En los problemas del 27 al 29, Hallar Un número \(\boldsymbol{M}\) que satisfaga la proposición dada.
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\[ \mid x + 2 \mid < 1 \Rightarrow \mid x^3 -x^2 + 2x + 1 \mid < M \]\[ \begin{aligned} &\mid x + 2 \mid < 1 \\[1em] &\hspace{2em} \Rightarrow \mid x^3 -x^2 + 2x + 1 \mid < M \end{aligned} \]
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\[ \mid x - 3 \mid < 1/2 \Rightarrow \frac{\mid x+2 \mid}{\mid x-2 \mid} < M \]\[ \begin{aligned} &\mid x - 3 \mid < 1/2 \\[1em] &\hspace{5em}\Rightarrow \frac{\mid x+2 \mid}{\mid x-2 \mid} < M \end{aligned} \]
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\[ \mid x - 1/4 \mid < 1/8 \Rightarrow \frac{\mid 16x+4 \mid}{1+x^2} < M \]\[ \begin{aligned} &\mid x - 1/4 \mid < 1/8 \\[1em] &\hspace{5em}\Rightarrow \frac{\mid 16x+4 \mid}{1+x^2} < M \end{aligned} \]
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Probar las siguientes expresiones:
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\[ \mid x - y \mid \geq \mid x \mid - \mid y \mid \]
Sugerencia: Aplicar la desigualdad triangular en: \(x = (x - y) + y\).
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\[ \mid x - y \mid \geq \mid y \mid - \mid x \mid \]
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\[ \left| \mid x \mid - \mid y \mid \right| \leq \mid x - y \mid \]
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Ver Respuestas
Respuestas
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\[ 9, \; 1 \]
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\[ -\frac{4}{3}, \; 2 \]
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\[ \frac{7}{2} \]
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\[ \frac{11}{4} \]
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\[ -1 \]
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\[ \frac{5}{3}, \, 3 \]
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\[ -6, \, 2 \]
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\[ \frac{1}{4}, \, 1 \]
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\[ -2, \, \frac{8}{3} \]
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\[ (1, \, 7) \]
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\[ \left( -\frac{16}{3}, \, \frac{14}{3} \right) \]
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\[ \left(-\frac{3}{2}, \, \frac{9}{2} \right) \]
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\[ \left[ -2, \, \frac{2}{3} \right] \]
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\[ \left( -\infty, \, -\frac{3}{5} \right] \cup \left[ -\frac{1}{5}, \, +\infty \right) \]
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\[ (-\infty, \, -1) \cup \left( -\frac{1}{2}, \, +\infty \right) \]
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\[ \left( -\infty, \, -\frac{5}{2} \right] \cup \left[ \frac{25}{2} +\infty \right) \]
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\[ [-\infty, \, -3] \cup [-1, \, 1] \cup [3, \, +\infty) \]
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\[ [-4, \, -1) \cup (1, \, 4] \]
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\[ (2, \, 4) – \{ 3 \} \]
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\[ \left( \frac{1}{2}, \, +\infty \right) \]
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\[ \left[\frac{2}{3}, \, 4 \right] \]
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\[ [-1, \, 2] – \left\lbrace \frac{1}{2} \right\rbrace \]
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\[ (-\infty, \, 1) \cup (2, \, +\infty) \]
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\[ [-2, \, 2] \]
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\[ (-1, \, 0) \cup (0, \, +\infty) \]
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\[ ( -\infty, \, -7 ] \cup \left[ \frac{1}{3}, \, +\infty \right) \]
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\[ M = 43 \]
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\[ M = 9 \]
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\[ M = 10 \]