## Tìm x, biết: a, $\frac{x-2}{x-6}$ < 0 b, $x^{2}$ + 4x > 0

Question

Tìm x, biết:
a, $\frac{x-2}{x-6}$ < 0 b, $x^{2}$ + 4x > 0

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1 year 2020-11-27T05:27:16+00:00 1 Answers 51 views 0

$\begin{array}{l} a)\dfrac{{x – 2}}{{x – 6}} < 0\\ \Rightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x – 2 > 0\\ x – 6 < 0 \end{array} \right.\\ \left\{ \begin{array}{l} x – 2 < 0\\ x – 6 > 0 \end{array} \right. \end{array} \right. \Rightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x > 2\\ x < 6 \end{array} \right.\\ \left\{ \begin{array}{l} x < 2\\ x > 6 \end{array} \right.\left( {ktm} \right) \end{array} \right.\\ \Rightarrow 2 < x < 6\\ \text{Vậy}\,2 < x < 6\\ b){x^2} + 4x > 0\\ \Rightarrow x\left( {x + 4} \right) > 0\\ \Rightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x > 0\\ x + 4 > 0 \end{array} \right.\\ \left\{ \begin{array}{l} x < 0\\ x + 4 < 0 \end{array} \right. \end{array} \right. \Rightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x > 0\\ x > – 4 \end{array} \right.\\ \left\{ \begin{array}{l} x < 0\\ x < – 4 \end{array} \right. \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l} x > 0\\ x < – 4 \end{array} \right.\\ \text{Vậy}\,x > 0\,\text{hoặc}\,x < – 4 \end{array}$