Problema 11

Obtener el conjunto solución de

  1. \(4x-5 < 7x-4\)
  2. \(x < \frac{5x-2}{2} < 7-x\)
  3. \(\frac{3}{x+5} < x\)
  4. \(x^{2} < 4x\)
  5. \(x^2 \leq x+2\)
  6. \(\frac{x}{1-x}\geq\frac{2+x}{x}\)
  7. \(\frac{x^2-1}{x^2+2} < 3\)
  8. \(\frac{x^2-4x}{x^2} \geq 1-x\)

P11.A

\(4x-5 < 7x-4\) \(\implies -5+4 < 7x-4x \implies -1 < 3x\implies x > -\frac{1}{3}\)
Por lo que \(\boxed{x\in(-\frac{1}{3},\infty)}\)

P11.B

\(x < \frac{5x-2}{2} < 7-x\)

  1. \(x < \frac{5x-2}{2} \implies 2x < 5x-2 \implies 2 < 3x \implies x > \frac{2}{3}\) Por lo que \(x\in(\frac{2}{3},\infty)\)
  2. \(\frac{5x-2}{2} < 7-x\implies 5x-2 < 14-2x \implies 7x < 16\implies x < \frac{16}{7}\)
    Por lo que \(x\in(-\infty,\frac{16}{7})\)
    3. Así, \(x\in(\frac{2}{3},\infty)\cap(-\infty,\frac{16}{7})\), de donde \(\boxed{x \in ( \frac{2}{3},\frac{16}{7} ) } \)

P11.C

\(\frac{3}{x+5} < x\)

  1. \(x+5 > 0\implies x > -5\)
    2. \(\implies 3 < x(x+5) = x^{2} + 5x\)
    \(\implies x^{2} + 5x + \frac{25}{4} > 3 + \frac{25}{4}\)
    \(\implies (x + \frac{5}{2})^{2} > \frac{37}{4}\)
    \(\implies |x + \frac{5}{2}| > \frac{\sqrt{37}}{2}\)
    \(\implies x + \frac{5}{2} > \frac{\sqrt{37}}{2}\) o \(x+\frac{5}{2} < -\frac{\sqrt{37}}{2}\)
    \(\implies x>\frac{-5+\sqrt{37}}{2}\) o \(x < \frac{-5-\sqrt{37}}{2}\)
    Así, \( x \in [( - \infty, \frac{-5 -\sqrt{37}}{2}) \cup (\frac{ -5 + \sqrt{37}}{2}, \infty)] \cap (-5,\infty)\).
    Por lo tanto, \(x\in(\frac{-5+\sqrt{37}}{2},\infty)\).

  2. \(x+5 < 0\implies x < -5\)
    3. \(\implies 3 < x(x+5)\) \(\vdots\) \( \mid x+\frac{5}{2} \mid < \frac{\sqrt{37}}{2}\) \(-\frac{\sqrt{37}}{2} < x + \frac{5}{2}<\frac{\sqrt{37}}{2}\) \(\frac{-5-\sqrt{37}}{2} < x < \frac{-5+\sqrt{37}}{2}\) Así, \(x\in(\frac{-5-\sqrt{37}}{2},\frac{-5+\sqrt{37}}{2})\cap(-\infty,-5)\). Por lo tanto, \(x\in(\frac{-5-\sqrt{37}}{2},-5)\). Así, \(\boxed{x\in(\frac{-5-\sqrt{37}}{2},-5) \cup (\frac{-5+\sqrt{37}}{2},\infty)}\)

P11.D

\(x^{2} < 4x\)

\(\implies x^{2} - 4x < 0 \implies x^{2} - 4x + 4 < 4 \implies (x-2)^{2} < 4\) \(\implies \lvert x-2 \rvert < 2\) \(\implies -2 < x-2 < 2\) \(\implies 0 < x < 4\) Por lo tanto, \(\boxed{x\in(0,4)}\)

P11.E

\(x^{2} \leq x+2\)

\(\implies x^{2} - x \leq 2 \implies x^{2} - x + \frac{1}{4} \leq 2 + \frac{1}{4} \implies (x - \frac{1}{2})^{2} \leq \frac{9}{4}\) \(\implies | x - \frac{1}{2} | \leq \frac{3}{2}\) \(\implies -\frac{3}{2} \leq x-\frac{1}{2} \leq \frac{3}{2}\) \(\implies -1 \leq x \leq 2 \) Por lo tanto, \(\boxed{x\in[-1,2]}\)

P11.F

\(\frac{x}{1-x} \geq \frac{2+x}{x}\)

  1. \(1-x > 0\) y \(x > 0 \implies x > 0\) y \(x < 1\implies 0 < x < 1\)
    2. \(\implies x^{2} \geq (2+x)(1-x) = 2 -2x + x -x^{2} \)
    \(\implies x^{2} \geq 2-x-x^2 \implies 2x^{2} + x \geq 2\)
    \(\implies x^{2} + \frac{1}{2}x \geq 1 \implies x^{2} + \frac{1}{2}x + \frac{1}{16} \geq 1 + \frac{1}{16}\)
    \(\implies (x + \frac{1}{4})^{2} \geq \frac{17}{16}\implies |x+\frac{1}{4}|\geq \frac{\sqrt{17}}{4}\)
    \(\implies x + \frac{1}{4}\geq\frac{\sqrt{17}}{4}\) o \(x+\frac{1}{4}\leq -\frac{\sqrt{17}}{4}\)
    \(\implies x \geq \frac{-1+\sqrt{17}}{4}\) o \(x \leq \frac{-1-\sqrt{17}}{4}\)
    Así, \(x\in\{(-\infty,\frac{-1-\sqrt{17}}{4})\cup(\frac{-1+\sqrt{17}}{4},\infty)\}\cap(0,1)\).
    Por lo tanto, \(x\in [\frac{-1+\sqrt{17}}{4},1)\).
  2. \(1-x > 0\) y \(x < \implies x < 0\) y \(x < 1\implies x < 0\)
    3. \(\implies x^2\leq (2+x)(1-x)\)
    \(\vdots\)
    \(\implies | x + \frac{1}{4} | \leq \frac{\sqrt{17}}{4}\)
    \(\implies -\frac{\sqrt{17}}{4} \leq x + \frac{1}{4} \leq \frac{\sqrt{17}}{4}\)
    \(\implies \frac{-1-\sqrt{17}}{4}\leq x \leq \frac{-1+\sqrt{17}}{4}\)
    Así, \(x\in [\frac{-1-\sqrt{17}}{4},\frac{-1+\sqrt{17}}{4}] \cap (-\infty,0)\).
    Por lo tanto, \(x \in [\frac{-1-\sqrt{17}}{4},0)\).
  3. \(1-x < 0\) y \(x > 0\implies x > 0\) y \(x > 1 \implies 1 < x\)
    4. \(\implies x^2\leq (2+x)(1-x)\)
    \(\vdots\)
    \(\implies \frac{-1-\sqrt{17}}{4} \leq x \leq \frac{-1+\sqrt{17}}{4}\)
    Así, \(x\in [\frac{-1-\sqrt{17}}{4},\frac{-1+\sqrt{17}}{4}]\cup(1,\infty)\).
    Por lo tanto, \(x \in \emptyset\).
  4. \(1-x < 0\) y \(x < 0\implies x < 0\) y \(x > 1\) !
    5. Por lo tanto, \(x\in\emptyset\).
    Así, \(\boxed{x \in [\frac{-1-\sqrt{17}}{4},0) \cup [\frac{-1+\sqrt{17}}{4},1)}\)

P11.G

\(\frac{x^{2} - 1}{x^{2} + 2} < 3\)

  1. \(x^{2} + 2 > 0 \implies x^{2} > -2\) (tautología, pues \(x^{2} \geq 0 > -2\))
    2. \(x^{2} - 1 = 3x^{2} + 6 \implies -7 < 2x^{2} \implies x^{2} > -\frac{7}{2}\) (tautología, pues \(x^{2} \geq 0 > -\frac{7}{2}\))
    Por lo tanto, \(x \in \mathbb{R}\).
  2. \(x^{2} + 2 < 0 \implies x^{2} < -2\) !
    3. Así, \(\boxed{x \in \mathbb{R}}\)

P11.H

\(\frac{x^{2} - 4x}{x^{2}} \geq 1-x\)

Como \(x^2\geq 0\),

\(x^{2} - 4x \geq x^{2}(1-x) \implies x^{2} - 4x \geq x^{2} - x^{3} \)
\(\implies -4x \geq -x^{3} \implies x^{3} - 4x \geq 0 \implies x(x^{2} - 4) \geq 0\)

Que se cumple si:
  1. \(x \geq 0\) y \(x^{2} - 4 \geq 0\) \(\implies x \geq 0\) y \(x^{2} \geq 4\)
    2. \(\implies x\geq 0\) y \((x\geq 2\) o \(x\leq -2)\)
    \(\implies x\geq 2\), i.e. \(x\in [2,\infty)\).
  2. \(x < 0\) y \(x^{2} - 4 < 0\) \(\implies x < 0\) y \(x^{2} < 4\)
    3. \(\implies x < 0\) y \(-2 < x < 2\)
    \(\implies 0 < x < 2\), i.e. \(x\in (-2,0)\).

Por lo tanto, \(x\in(-2,0) \cup [2,\infty)\)