CALCULO I - CALCULO NIVELI 6- Cuando dos resistencias \( R_{1} \) y \( R_{2} \) están conectadas en paralelo, la resistencia total \( R \) satisface la ecuación \( \frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}} \). Encuentre \( R_{1} \) para un circuito en paralelo en el que \( R_{2}=4[ \) ohm \( ] \) y \( R \) debe ser al menos de \( 2,5[ \) ohm \( ] \). 7- Encuentre y grafique en la recta real el conjunto solución de \( \begin{array}{lll}\text { a) }|4+7 x|=5 & \text { b) }|x-8|=\mid 5 x-16 & \text { e) }\left|x+\frac{2}{3}\right|=|-2 x+3| \\ \text { d) } 23-|8 x|=7-x^{2} & \text { e) }|9-x| \leq \sqrt{5} & \text { f) } 12-|x+5|<0 \\ \text { g) } 49-7 x^{2} \leq 0 & \text { h) }-1+\left(x+\frac{5}{3}\right)^{2} \leq 2 & \text { i) } \frac{x+9}{|x-5| x^{5}+0}<\end{array} \)

Solve the inequality by following steps: - step0: Solve for \(x\): \(12-\left|x+5\right|<0\) - step1: Rearrange the terms: \(-\left|x+5\right|<-12\) - step2: Calculate: \(\left|x+5\right|>12\) - step3: Separate into possible cases: \(\begin{align}&x+5>12\\&x+5<-12\end{align}\) - step4: Solve the inequality: \(\begin{align}&x>7\\&x<-17\end{align}\) - step5: Find the union: \(x \in \left(-\infty,-17\right)\cup \left(7,+\infty\right)\) Solve the equation \( |9-x| \leq \sqrt{5} \). Solve the inequality by following steps: - step0: Solve for \(x\): \(\left|9-x\right|\leq \sqrt{5}\) - step1: Separate into possible cases: \(\left\{ \begin{array}{l}9-x\leq \sqrt{5}\\9-x\geq -\sqrt{5}\end{array}\right.\) - step2: Solve the inequality: \(\left\{ \begin{array}{l}x\geq -\sqrt{5}+9\\x\leq \sqrt{5}+9\end{array}\right.\) - step3: Find the intersection: \(-\sqrt{5}+9\leq x\leq \sqrt{5}+9\) Solve the equation \( |4+7x|=5 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\left|4+7x\right|=5\) - step1: Separate into possible cases: \(\begin{align}&4+7x=5\\&4+7x=-5\end{align}\) - step2: Solve the equation: \(\begin{align}&x=\frac{1}{7}\\&x=-\frac{9}{7}\end{align}\) - step3: Rewrite: \(x_{1}=-\frac{9}{7},x_{2}=\frac{1}{7}\) Solve the equation \( 49-7x^{2} \leq 0 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(49-7x^{2}\leq 0\) - step1: Rewrite the expression: \(49-7x^{2}=0\) - step2: Move the constant to the right side: \(-7x^{2}=0-49\) - step3: Remove 0: \(-7x^{2}=-49\) - step4: Change the signs: \(7x^{2}=49\) - step5: Divide both sides: \(\frac{7x^{2}}{7}=\frac{49}{7}\) - step6: Divide the numbers: \(x^{2}=7\) - step7: Simplify the expression: \(x=\pm \sqrt{7}\) - step8: Separate into possible cases: \(\begin{align}&x=\sqrt{7}\\&x=-\sqrt{7}\end{align}\) - step9: Determine the test intervals: \(\begin{align}&x<-\sqrt{7}\\&-\sqrt{7}\sqrt{7}\end{align}\) - step10: Choose a value: \(\begin{align}&x_{1}=-4\\&x_{2}=0\\&x_{3}=4\end{align}\) - step11: Test the chosen value: \(\begin{align}&x<-\sqrt{7}\textrm{ }\textrm{is the solution}\\&-\sqrt{7}\sqrt{7}\textrm{ }\textrm{is the solution}\end{align}\) - step12: Include the critical value: \(\begin{align}&x\leq -\sqrt{7}\textrm{ }\textrm{is the solution}\\&x\geq \sqrt{7}\textrm{ }\textrm{is the solution}\end{align}\) - step13: The final solution is \(x \in \left(-\infty,-\sqrt{7}\right]\cup \left[\sqrt{7},+\infty\right):\) \(x \in \left(-\infty,-\sqrt{7}\right]\cup \left[\sqrt{7},+\infty\right)\) Solve the equation \( |x-8|=|5x-16| \). Solve the equation by following steps: - step0: Solve for \(x\): \(\left|x-8\right|=\left|5x-16\right|\) - step1: Evaluate: \(\begin{align}&x-8=5x-16\\&x-8=-\left(5x-16\right)\end{align}\) - step2: Calculate: \(\begin{align}&x=2\\&x=4\end{align}\) - step3: Rewrite: \(x_{1}=2,x_{2}=4\) Solve the equation \( 23-|8x|=7-x^{2} \). Solve the equation by following steps: - step0: Solve for \(x\): \(23-\left|8x\right|=7-x^{2}\) - step1: Move the expression to the left side: \(23-\left|8x\right|-\left(7-x^{2}\right)=0\) - step2: Subtract the terms: \(16-\left|8x\right|+x^{2}=0\) - step3: Separate into possible cases: \(\begin{align}&16-8x+x^{2}=0,8x\geq 0\\&16-\left(-8x\right)+x^{2}=0,8x<0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=4,x\geq 0\\&x=-4,x<0\end{align}\) - step5: Find the intersection: \(\begin{align}&x=4\\&x=-4\end{align}\) - step6: Rewrite: \(x_{1}=-4,x_{2}=4\) Solve the equation \( \left|x+\frac{2}{3}\right|=|-2x+3| \). Solve the equation by following steps: - step0: Solve for \(x\): \(\left|x+\frac{2}{3}\right|=\left|-2x+3\right|\) - step1: Rewrite the expression: \(\left|x+\frac{2}{3}\right|=\left|2x-3\right|\) - step2: Evaluate: \(\begin{align}&x+\frac{2}{3}=2x-3\\&x+\frac{2}{3}=-\left(2x-3\right)\end{align}\) - step3: Calculate: \(\begin{align}&x=\frac{11}{3}\\&x=\frac{7}{9}\end{align}\) - step4: Rewrite: \(x_{1}=\frac{7}{9},x_{2}=\frac{11}{3}\) Solve the equation \( \frac{x+9}{|x-5| x^{5}+0}<0 \). Solve the inequality by following steps: - step0: Solve the inequality by separating into cases: \(\frac{x+9}{\left|x-5\right|\times x^{5}+0}<0\) - step1: Find the domain: \(\frac{x+9}{\left|x-5\right|\times x^{5}+0}<0,x \in \left(-\infty,0\right)\cup \left(0,5\right)\cup \left(5,+\infty\right)\) - step2: Simplify: \(\frac{x+9}{x^{5}\left|x-5\right|}<0\) - step3: Separate into possible cases: \(\begin{align}&\left\{ \begin{array}{l}x+9>0\\x^{5}\left|x-5\right|<0\end{array}\right.\\&\left\{ \begin{array}{l}x+9<0\\x^{5}\left|x-5\right|>0\end{array}\right.\end{align}\) - step4: Solve the inequality: \(\begin{align}&\left\{ \begin{array}{l}x>-9\\x<0\end{array}\right.\\&\left\{ \begin{array}{l}x<-9\\x \in \left(0,5\right)\cup \left(5,+\infty\right)\end{array}\right.\end{align}\) - step5: Find the intersection: \(\begin{align}&-9\frac{3\sqrt{3}-5}{3}\end{align}\) - step11: Choose a value: \(\begin{align}&x_{1}=-4\\&x_{2}=-2\\&x_{3}=1\end{align}\) - step12: Test the chosen value: \(\begin{align}&x<-\frac{3\sqrt{3}+5}{3}\textrm{ }\textrm{is not a solution}\\&-\frac{3\sqrt{3}+5}{3}\frac{3\sqrt{3}-5}{3}\textrm{ }\textrm{is not a solution}\end{align}\) - step13: Include the critical value: \(\begin{align}&-\frac{3\sqrt{3}+5}{3}\leq x\leq \frac{3\sqrt{3}-5}{3}\textrm{ }\textrm{is the solution}\end{align}\) - step14: The final solution is \(-\frac{3\sqrt{3}+5}{3}\leq x\leq \frac{3\sqrt{3}-5}{3}:\) \(-\frac{3\sqrt{3}+5}{3}\leq x\leq \frac{3\sqrt{3}-5}{3}\) ### Problema 6: Resistencia en Paralelo La ecuación que relaciona las resistencias en paralelo es: \[ \frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} \] Dado que \( R_{2} = 4 \) ohmios y \( R \) debe ser al menos \( 2.5 \) ohmios, podemos sustituir \( R \) en la ecuación: \[ \frac{1}{2.5} = \frac{1}{R_{1}} + \frac{1}{4} \] Ahora, resolvemos para \( R_{1} \): \[ \frac{1}{R_{1}} = \frac{1}{2.5} - \frac{1}{4} \] Calculamos \( \frac{1}{2.5} \) y \( \frac{1}{4} \): \[ \frac{1}{2.5} = 0.4 \quad \text{y} \quad \frac{1}{4} = 0.25 \] Por lo tanto: \[ \frac{1}{R_{1}} = 0.4 - 0.25 = 0.15 \] Finalmente, invertimos para encontrar \( R_{1} \): \[ R_{1} = \frac{1}{0.15} \approx 6.67 \text{ ohmios} \] ### Problema 7: Conjuntos Solución a) Para \( |4 + 7x| = 5 \): \[ x_{1} = -\frac{9}{7}, \quad x_{2} = \frac{1}{7} \] b) Para \( |x - 8| = |5x - 16| \): \[ x_{1} = 2, \quad x_{2} = 4 \] c) Para \( \left|x + \frac{2}{3}\right| = |-2x + 3| \): \[ x_{1} = \frac{7}{9}, \quad x_{2} = \frac{11}{3} \] d) Para \( 23 - |8x| = 7 - x^{2} \): \[ x_{1} = -4, \quad x_{2} = 4 \] e) Para \( |9 - x| \leq \sqrt{5} \): \[ -\sqrt{5} + 9 \leq x \leq \sqrt{5} + 9 \] f) Para \( 12 - |x + 5| < 0 \): \[ x \in (-\infty, -17) \cup (7, +\infty) \] g) Para \( 49 - 7x^{2} \leq 0 \): \[ x \in (-\infty, -\sqrt{7}] \cup [\sqrt{7}, +\infty) \] h) Para \( -1 + \left(x + \frac{5}{3}\right)^{2} \leq 2 \): \[ -\frac{3\sqrt{3} + 5}{3} \leq x \leq \frac{3\sqrt{3} - 5}{3} \] i) Para \( \frac{x + 9}{|x - 5| x^{5} + 0} < 0 \): \[ -9 < x < 0 \] ### Gráficas Para graficar los conjuntos solución, se pueden representar en la recta real los intervalos y puntos encontrados. Si necesitas las gráficas específicas, por favor indícalo.