Vodka 989 Questions 2k Answers 0 Best Answers 22 Points View Profile0 Vodka Asked: Tháng Mười Một 27, 20202020-11-27T05:27:16+00:00 2020-11-27T05:27:16+00:00In: Môn ToánTìm x, biết: a, $\frac{x-2}{x-6}$ < 0 b, $x^{2}$ + 4x > 00Tìm x, biết: a, $\frac{x-2}{x-6}$ < 0 b, $x^{2}$ + 4x > 0 ShareFacebookRelated Questions GIÚP IEM VỚI THÁNKS Ca dao Hưng Yên nói về tình cảm gia đình Cách soạn bài ngữ văn lớp 8 văn bản cô bé bán diêm 1 AnswerOldestVotedRecentTryphena 933 Questions 2k Answers 0 Best Answers 7 Points View Profile Tryphena 2020-11-27T05:29:02+00:00Added an answer on Tháng Mười Một 27, 2020 at 5:29 sáng 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}$0Reply Share ShareShare on FacebookLeave an answerLeave an answerHủy By answering, you agree to the Terms of Service and Privacy Policy .* Lưu tên của tôi, email, và trang web trong trình duyệt này cho lần bình luận kế tiếp của tôi.
Tryphena
Đáp án:
$\begin{array}{l}
a)\dfrac{{x – 2}}{{x – 6}} < 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x – 2 > 0\\
x – 6 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x – 2 < 0\\
x – 6 > 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 2\\
x < 6
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 2\\
x > 6
\end{array} \right.\left( {ktm} \right)
\end{array} \right.\\
\Rightarrow 2 < x < 6\\
\text{Vậy}\,2 < x < 6\\
b){x^2} + 4x > 0\\
\Rightarrow x\left( {x + 4} \right) > 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 0\\
x + 4 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 0\\
x + 4 < 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 0\\
x > – 4
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 0\\
x < – 4
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x > 0\\
x < – 4
\end{array} \right.\\
\text{Vậy}\,x > 0\,\text{hoặc}\,x < – 4
\end{array}$