Toán Tìm x, biết: a, $\frac{x-2}{x-6}$ < 0 b, $x^{2}$ + 4x > 0 19/07/2021 By Jade Tìm x, biết: a, $\frac{x-2}{x-6}$ < 0 b, $x^{2}$ + 4x > 0
Đáp án: $\begin{array}{l}a)\dfrac{{x – 2}}{{x – 6}} < 0\\ \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x – 2 > 0\\x – 6 < 0\end{array} \right.\\\left\{ \begin{array}{l}x – 2 < 0\\x – 6 > 0\end{array} \right.\end{array} \right. \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x > 2\\x < 6\end{array} \right.\\\left\{ \begin{array}{l}x < 2\\x > 6\end{array} \right.\left( {ktm} \right)\end{array} \right.\\ \Rightarrow 2 < x < 6\\\text{Vậy}\,2 < x < 6\\b){x^2} + 4x > 0\\ \Rightarrow x\left( {x + 4} \right) > 0\\ \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x > 0\\x + 4 > 0\end{array} \right.\\\left\{ \begin{array}{l}x < 0\\x + 4 < 0\end{array} \right.\end{array} \right. \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x > 0\\x > – 4\end{array} \right.\\\left\{ \begin{array}{l}x < 0\\x < – 4\end{array} \right.\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}x > 0\\x < – 4\end{array} \right.\\\text{Vậy}\,x > 0\,\text{hoặc}\,x < – 4\end{array}$ Trả lời