1. Halla el conjunto de solución de lás sigulentes I necuaciunes cuadrúticas: a) $$ x^{2}+6 x+50 $$ b) $$ 4 x^{2}-25 \leqslant 0 $$ c) $$ x^{2}-11 x0 $$ e) $$ x^{2}-x+60 $$ f) $$ 15 x^{2}-2 \geqslant 7 x $$ h) $$ 7 x^{2}+21 x-280 $$ c) $$ (x+2) \cdot(x+3)

Question

1. Halla el conjunto de solución de lás sigulentes I necuaciunes cuadrúticas: a) $$ x^{2}+6 x+5>0 $$ b) $$ 4 x^{2}-25 \leqslant 0 $$ c) $$ x^{2}-11 x<-18 $$ d) $$ x^{2}-8 x-1008>0 $$ e) $$ x^{2}-x+6>0 $$ f) $$ 15 x^{2}-2 \geqslant 7 x $$ h) $$ 7 x^{2}+21 x-28<0 $$ 2. $$ H $$ alla el conjunto de solución: a) $$ (x+2) \cdot(x-5) \leq 0 $$ b) $$ (x-3) \cdot(x-1)>0 $$ c) $$ (x+2) \cdot(x+3)<0 $$ d) $$ (x+9) \cdot(x-10) \geqslant 0 $$ e) $$ \frac{(x+4)}{(x-5)}<0 $$

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Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: $$\frac{\left(x+4 ight)}{\left(x-5 ight)}<0$$ - step1: Find the domain: $$\frac{\left(x+4 ight)}{\left(x-5 ight)}<0,x eq 5$$ - step2: Remove the parentheses: $$\frac{x+4}{x-5}<0$$ - step3: Set the numerator and denominator of $$\frac{x+4}{x-5}$$ equal to 0$$:$$ $$\begin{align}&x+4=0\&x-5=0\end{align}$$ - step4: Calculate: $$\begin{align}&x=-4\&x=5\end{align}$$ - step5: Determine the test intervals: $$\begin{align}&x<-4\&-45\end{align}$$ - step6: Choose a value: $$\begin{align}&x_{1}=-5\&x_{2}=1\&x_{3}=6\end{align}$$ - step7: Test the chosen value: $$\begin{align}&x<-4 extrm{ } extrm{is not a solution}\&-45 extrm{ } extrm{is not a solution}\end{align}$$ - step8: The final solution is $$-40 $$. Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: $$\left(x-3 ight)\left(x-1 ight)>0$$ - step1: Rewrite the expression: $$\left(x-3 ight)\left(x-1 ight)=0$$ - step2: Separate into possible cases: $$\begin{align}&x-3=0\&x-1=0\end{align}$$ - step3: Solve the equation: $$\begin{align}&x=3\&x=1\end{align}$$ - step4: Determine the test intervals: $$\begin{align}&x<1\&13\end{align}$$ - step5: Choose a value: $$\begin{align}&x_{1}=0\&x_{2}=2\&x_{3}=4\end{align}$$ - step6: Test the chosen value: $$\begin{align}&x<1 extrm{ } extrm{is the solution}\&13 extrm{ } extrm{is the solution}\end{align}$$ - step7: The final solution is $$x \in \left(-\infty,1 ight)\cup \left(3,+\infty ight):$$ $$x \in \left(-\infty,1 ight)\cup \left(3,+\infty ight)$$ Solve the equation $$ x^{2}-11 x<-18 $$. Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: $$x^{2}-11x<-18$$ - step1: Move the expression to the left side: $$x^{2}-11x-\left(-18 ight)<0$$ - step2: Remove the parentheses: $$x^{2}-11x+18<0$$ - step3: Rewrite the expression: $$x^{2}-11x+18=0$$ - step4: Factor the expression: $$\left(x-9 ight)\left(x-2 ight)=0$$ - step5: Separate into possible cases: $$\begin{align}&x-9=0\&x-2=0\end{align}$$ - step6: Solve the equation: $$\begin{align}&x=9\&x=2\end{align}$$ - step7: Determine the test intervals: $$\begin{align}&x<2\&29\end{align}$$ - step8: Choose a value: $$\begin{align}&x_{1}=1\&x_{2}=6\&x_{3}=10\end{align}$$ - step9: Test the chosen value: $$\begin{align}&x<2 extrm{ } extrm{is not a solution}\&29 extrm{ } extrm{is not a solution}\end{align}$$ - step10: The final solution is $$20 $$. Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: $$x^{2}+6x+5>0$$ - step1: Rewrite the expression: $$x^{2}+6x+5=0$$ - step2: Factor the expression: $$\left(x+1 ight)\left(x+5 ight)=0$$ - step3: Separate into possible cases: $$\begin{align}&x+1=0\&x+5=0\end{align}$$ - step4: Solve the equation: $$\begin{align}&x=-1\&x=-5\end{align}$$ - step5: Determine the test intervals: $$\begin{align}&x<-5\&-5-1\end{align}$$ - step6: Choose a value: $$\begin{align}&x_{1}=-6\&x_{2}=-3\&x_{3}=0\end{align}$$ - step7: Test the chosen value: $$\begin{align}&x<-5 extrm{ } extrm{is the solution}\&-5-1 extrm{ } extrm{is the solution}\end{align}$$ - step8: The final solution is $$x \in \left(-\infty,-5 ight)\cup \left(-1,+\infty ight):$$ $$x \in \left(-\infty,-5 ight)\cup \left(-1,+\infty ight)$$ Solve the equation $$ (x+2) \cdot(x-5) \leq 0 $$. Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: $$\left(x+2 ight)\left(x-5 ight)\leq 0$$ - step1: Rewrite the expression: $$\left(x+2 ight)\left(x-5 ight)=0$$ - step2: Separate into possible cases: $$\begin{align}&x+2=0\&x-5=0\end{align}$$ - step3: Solve the equation: $$\begin{align}&x=-2\&x=5\end{align}$$ - step4: Determine the test intervals: $$\begin{align}&x<-2\&-25\end{align}$$ - step5: Choose a value: $$\begin{align}&x_{1}=-3\&x_{2}=2\&x_{3}=6\end{align}$$ - step6: Test the chosen value: $$\begin{align}&x<-2 extrm{ } extrm{is not a solution}\&-25 extrm{ } extrm{is not a solution}\end{align}$$ - step7: Include the critical value: $$\begin{align}&-2\leq x\leq 5 extrm{ } extrm{is the solution}\end{align}$$ - step8: The final solution is $$-2\leq x\leq 5:$$ $$-2\leq x\leq 5$$ Solve the equation $$ 7 x^{2}+21 x-28<0 $$. Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: $$7x^{2}+21x-28<0$$ - step1: Rewrite the expression: $$7x^{2}+21x-28=0$$ - step2: Factor the expression: $$7\left(x-1 ight)\left(x+4 ight)=0$$ - step3: Divide the terms: $$\left(x-1 ight)\left(x+4 ight)=0$$ - step4: Separate into possible cases: $$\begin{align}&x-1=0\&x+4=0\end{align}$$ - step5: Solve the equation: $$\begin{align}&x=1\&x=-4\end{align}$$ - step6: Determine the test intervals: $$\begin{align}&x<-4\&-41\end{align}$$ - step7: Choose a value: $$\begin{align}&x_{1}=-5\&x_{2}=-2\&x_{3}=2\end{align}$$ - step8: Test the chosen value: $$\begin{align}&x<-4 extrm{ } extrm{is not a solution}\&-41 extrm{ } extrm{is not a solution}\end{align}$$ - step9: The final solution is $$-4-2\end{align}$$ - step5: Choose a value: $$\begin{align}&x_{1}=-4\&x_{2}=-\frac{5}{2}\&x_{3}=-1\end{align}$$ - step6: Test the chosen value: $$\begin{align}&x<-3 extrm{ } extrm{is not a solution}\&-3-2 extrm{ } extrm{is not a solution}\end{align}$$ - step7: The final solution is $$-30 $$. Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: $$x^{2}-x+6>0$$ - step1: Rewrite the expression: $$x^{2}-x+6=0$$ - step2: Add or subtract both sides: $$x^{2}-x=-6$$ - step3: Add the same value to both sides: $$x^{2}-x+\frac{1}{4}=-6+\frac{1}{4}$$ - step4: Simplify the expression: $$\left(x-\frac{1}{2} ight)^{2}=-\frac{23}{4}$$ - step5: The statement is false for any value of $$x:$$ $$x otin \mathbb{R}$$ - step6: Choose a value: $$x=0$$ - step7: Test the chosen value: $$x \in \mathbb{R}$$ Solve the equation $$ x^{2}-8 x-1008>0 $$. Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: $$x^{2}-8x-1008>0$$ - step1: Rewrite the expression: $$x^{2}-8x-1008=0$$ - step2: Factor the expression: $$\left(x-36 ight)\left(x+28 ight)=0$$ - step3: Separate into possible cases: $$\begin{align}&x-36=0\&x+28=0\end{align}$$ - step4: Solve the equation: $$\begin{align}&x=36\&x=-28\end{align}$$ - step5: Determine the test intervals: $$\begin{align}&x<-28\&-2836\end{align}$$ - step6: Choose a value: $$\begin{align}&x_{1}=-29\&x_{2}=4\&x_{3}=37\end{align}$$ - step7: Test the chosen value: $$\begin{align}&x<-28 extrm{ } extrm{is the solution}\&-2836 extrm{ } extrm{is the solution}\end{align}$$ - step8: The final solution is $$x \in \left(-\infty,-28 ight)\cup \left(36,+\infty ight):$$ $$x \in \left(-\infty,-28 ight)\cup \left(36,+\infty ight)$$ Aquí están los conjuntos de solución para las inecuaciones cuadráticas y los productos de factores que proporcionaste: ### 1. Inecuaciones cuadráticas: a) $$ x^{2}+6 x+5>0 $$ **Solución:** $$ x \in (-\infty,-5) \cup (-1,+\infty) $$ b) $$ 4 x^{2}-25 \leqslant 0 $$ **Solución:** No se pudo resolver debido a un error en la sintaxis de la expresión. c) $$ x^{2}-11 x<-18 $$ **Solución:** $$ 2 < x < 9 $$ d) $$ x^{2}-8 x-1008>0 $$ **Solución:** $$ x \in (-\infty,-28) \cup (36,+\infty) $$ e) $$ x^{2}-x+6>0 $$ **Solución:** $$ x \in \mathbb{R} $$ (todas las soluciones reales) f) $$ 15 x^{2}-2 \geqslant 7 x $$ **Solución:** No se pudo resolver debido a un error en la sintaxis de la expresión. g) $$ 7 x^{2}+21 x-28<0 $$ **Solución:** $$ -4 < x < 1 $$ ### 2. Conjuntos de solución: a) $$ (x+2) \cdot(x-5) \leq 0 $$ **Solución:** $$ -2 \leq x \leq 5 $$ b) $$ (x-3) \cdot(x-1)>0 $$ **Solución:** $$ x \in (-\infty,1) \cup (3,+\infty) $$ c) $$ (x+2) \cdot(x+3)<0 $$ **Solución:** $$ -3 < x < -2 $$ d) $$ (x+9) \cdot(x-10) \geqslant 0 $$ **Solución:** No se pudo resolver debido a un error en la sintaxis de la expresión. e) $$ \frac{(x+4)}{(x-5)}<0 $$ **Solución:** $$ -4 < x < 5 $$ Si necesitas más detalles o aclaraciones sobre alguna de las soluciones, no dudes en preguntar. Aquí están las soluciones para las inecuaciones y los productos de factores: 1. **Inecuaciones cuadráticas:** - a) $$ x^{2}+6 x+5>0 $$: $$ x < -5 $$ o $$ x > -1 $$ - b) $$ 4 x^{2}-25 \leqslant 0 $$: No se pudo resolver. - c) $$ x^{2}-11 x<-18 $$: $$ 2 < x < 9 $$ - d) $$ x^{2}-8 x-1008>0 $$: $$ x < -28 $$ o $$ x > 36 $$ - e) $$ x^{2}-x+6>0 $$: Todas las soluciones reales - f) $$ 15 x^{2}-2 \geqslant 7 x $$: No se pudo resolver. - g) $$ 7 x^{2}+21 x-28<0 $$: $$ -4 < x < 1 $$ 2. **Conjuntos de solución:** - a) $$ (x+2) \cdot(x-5) \leq 0 $$: $$ -2 \leq x \leq 5 $$ - b) $$ (x-3) \cdot(x-1)>0 $$: $$ x < 1 $$ o $$ x > 3 $$ - c) $$ (x+2) \cdot(x+3)<0 $$: $$ -3 < x < -2 $$ - d) $$ (x+9) \cdot(x-10) \geqslant 0 $$: No se pudo resolver. - e) $$ \frac{(x+4)}{(x-5)}<0 $$: $$ -4 < x < 5 $$ Si necesitas más detalles o aclaraciones, pregunta.

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