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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\[ \mid x – 1/4 \mid < 1/8 \Rightarrow \frac{\mid 16x+4 \mid}{1+x^2} < M \]
\[ \begin{aligned} &\mid x – 1/4 \mid < 1/8 \\[1em] &\hspace{5em}\Rightarrow \frac{\mid 16x+4 \mid}{1+x^2} < M \end{aligned} \]
- 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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\[ \left| \mid x \mid – \mid y \mid \right| \leq \mid x – y \mid \]