giải bất phương trình $4x^{2}$ – 12x $\leq$ $\frac{-135}{16}$ $x^{2}$ – x $\leq$ 2 04/07/2021 Bởi Reagan giải bất phương trình $4x^{2}$ – 12x $\leq$ $\frac{-135}{16}$ $x^{2}$ – x $\leq$ 2
Đáp án: a) \(\left\{ \begin{array}{l}x \le \dfrac{{15}}{8}\\x \ge \dfrac{9}{8}\end{array} \right.\) b) \(\left\{ \begin{array}{l}x \le 2\\x \ge – 1\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}a)4{x^2} – 12x \le – \dfrac{{135}}{{16}}\\ \to 4{x^2} – 12x + 9 \le \dfrac{9}{{16}}\\ \to {\left( {2x – 3} \right)^2} \le \dfrac{9}{{16}}\\ \to \left\{ \begin{array}{l}2x – 3 \le \dfrac{3}{4}\\2x – 3 \ge – \dfrac{3}{4}\end{array} \right.\\ \to \left\{ \begin{array}{l}x \le \dfrac{{15}}{8}\\x \ge \dfrac{9}{8}\end{array} \right.\\b){x^2} – x \le 2\\ \to {x^2} – x – 2 \le 0\\ \to \left( {x – 2} \right)\left( {x + 1} \right) \le 0\\ \to \left[ \begin{array}{l}\left\{ \begin{array}{l}x – 2 \ge 0\\x + 1 \le 0\end{array} \right.\\\left\{ \begin{array}{l}x – 2 \le 0\\x + 1 \ge 0\end{array} \right.\end{array} \right. \to \left[ \begin{array}{l}\left\{ \begin{array}{l}x \ge 2\\x \le – 1\end{array} \right.\left( l \right)\\\left\{ \begin{array}{l}x \le 2\\x \ge – 1\end{array} \right.\end{array} \right.\\ \to \left\{ \begin{array}{l}x \le 2\\x \ge – 1\end{array} \right.\end{array}\) Bình luận
Đáp án:
a) \(\left\{ \begin{array}{l}
x \le \dfrac{{15}}{8}\\
x \ge \dfrac{9}{8}
\end{array} \right.\)
b) \(\left\{ \begin{array}{l}
x \le 2\\
x \ge – 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)4{x^2} – 12x \le – \dfrac{{135}}{{16}}\\
\to 4{x^2} – 12x + 9 \le \dfrac{9}{{16}}\\
\to {\left( {2x – 3} \right)^2} \le \dfrac{9}{{16}}\\
\to \left\{ \begin{array}{l}
2x – 3 \le \dfrac{3}{4}\\
2x – 3 \ge – \dfrac{3}{4}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \le \dfrac{{15}}{8}\\
x \ge \dfrac{9}{8}
\end{array} \right.\\
b){x^2} – x \le 2\\
\to {x^2} – x – 2 \le 0\\
\to \left( {x – 2} \right)\left( {x + 1} \right) \le 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x – 2 \ge 0\\
x + 1 \le 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x – 2 \le 0\\
x + 1 \ge 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 2\\
x \le – 1
\end{array} \right.\left( l \right)\\
\left\{ \begin{array}{l}
x \le 2\\
x \ge – 1
\end{array} \right.
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \le 2\\
x \ge – 1
\end{array} \right.
\end{array}\)