Giải bất phương trình và biểu diễn nghiệm trên trục số `c) 12x^2 − 25x + 12 < 0` `d) x^2 + 3x − 1 ≤ 2x + 5`
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Đáp án+Giải thích các bước giải:
$c)12x^2-25x+12<0\\ \Leftrightarrow 12x^2-16-9x+12 <0\\ \Leftrightarrow 4x(3x-4)-3(3x-4) <0\\ \Leftrightarrow (4x-3)(3x-4)<0\\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} 4x-3 >0 \\3x-4<0 \end{array} \right. \\ \left\{\begin{array}{l} 4x-3 <0 \\3x-4>0 \end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} 4x>3 \\3x<4 \end{array} \right. \\ \left\{\begin{array}{l} 4x<3 \\3x>4 \end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} x>\dfrac{3}{4} \\x<\dfrac{4}{3} \end{array} \right. \\ \left\{\begin{array}{l} x<\dfrac{3}{4} \\x>\dfrac{4}{3} \end{array} \right. (L)\end{array} \right.\\ \Leftrightarrow\left\{\begin{array}{l} x>\dfrac{3}{4} \\x<\dfrac{4}{3} \end{array} \right.\\ \Leftrightarrow \dfrac{3}{4}<x<\dfrac{4}{3}\\ d)x^2+3x-1 \le 2x+5\\ \Leftrightarrow x^2+x-6 \le 0\\ \Leftrightarrow x^2-2x+3x-6 \le 0\\ \Leftrightarrow x(x-2)+3(x-2) \le 0\\ \Leftrightarrow (x+3)(x-2) \le 0\\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} x+3 \ge 0 \\ x-2 \le 0 \end{array} \right. \\ \left\{\begin{array}{l} x+3 \le 0 \\ x-2 \ge 0 \end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} x \ge -3 \\ x \le 2 \end{array} \right. \\ \left\{\begin{array}{l} x \le -3 \\ x \ge 2 \end{array} \right. (L)\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge -3 \\ x \le 2 \end{array} \right.\\ \Leftrightarrow -3 \le x \le 2.$