Giải bất phương trình bậc nhất. (4-x)(2x+3y)≤0 24/07/2021 Bởi Abigail Giải bất phương trình bậc nhất. (4-x)(2x+3y)≤0
Đáp án: $\begin{array}{l}\left( {4 – x} \right).\left( {2x + 3y} \right) \le 0\\ \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}4 – x \le 0\\2x + 3y \ge 0\end{array} \right.\\\left\{ \begin{array}{l}4 – x \ge 0\\2x + 3y \le 0\end{array} \right.\end{array} \right. \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x \ge 4\\x \ge – \frac{{3y}}{2}\end{array} \right.\\\left\{ \begin{array}{l}x \le 4\\x \le – \frac{{3y}}{2}\end{array} \right.\end{array} \right. \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x \ge 4\\4 \ge – \frac{{3y}}{2} \Rightarrow y \ge – \frac{8}{3}\end{array} \right.\\\left\{ \begin{array}{l}x \le 4\\4 \le – \frac{{3y}}{2} \Rightarrow y \ge – \frac{8}{3}\end{array} \right.\end{array} \right.\\Vậy\left\{ \begin{array}{l}x \ge 4\\y \ge – \frac{8}{3}\end{array} \right.\,hoac\,\,\left\{ \begin{array}{l}x \le 4\\y \le – \frac{8}{3}\end{array} \right.\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
\left( {4 – x} \right).\left( {2x + 3y} \right) \le 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
4 – x \le 0\\
2x + 3y \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
4 – x \ge 0\\
2x + 3y \le 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 4\\
x \ge – \frac{{3y}}{2}
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le 4\\
x \le – \frac{{3y}}{2}
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 4\\
4 \ge – \frac{{3y}}{2} \Rightarrow y \ge – \frac{8}{3}
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le 4\\
4 \le – \frac{{3y}}{2} \Rightarrow y \ge – \frac{8}{3}
\end{array} \right.
\end{array} \right.\\
Vậy\left\{ \begin{array}{l}
x \ge 4\\
y \ge – \frac{8}{3}
\end{array} \right.\,hoac\,\,\left\{ \begin{array}{l}
x \le 4\\
y \le – \frac{8}{3}
\end{array} \right.
\end{array}$