Đáp án:

$a)\ S = \left\{0; \dfrac{3}{2}\right\}$

$b)\ S = \{-2\}$

$c)\ S = \{x \mid x < 4\}$

$d)\ S = \{x \mid x \ge -7\}$

Giải thích:

$a)\ 9x^2(2x - 3) = 0$

$\Leftrightarrow \left[\begin{array}{l} 9x^2 = 0 \\ 2x - 3 = 0 \end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l} x = 0 \\ 2x = 3 \end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l} x = 0 \\ x = \dfrac{3}{2} \end{array}\right.$

Vậy $S = \left\{0; \dfrac{3}{2}\right\}$

$b)\ \dfrac{x + 3}{x - 3} = \dfrac{3}{x^2 - 3x} + \dfrac{1}{x}$

ĐKXĐ: $x \ne 0; x \ne 3$

$\Leftrightarrow \dfrac{x(x + 3)}{x(x - 3)} = \dfrac{3}{x(x - 3)} + \dfrac{x - 3}{x(x - 3)}$

$\Rightarrow x(x + 3) = 3 + x - 3$

$\Leftrightarrow x^2 + 3x = x$

$\Leftrightarrow x^2 + 2x = 0$

$\Leftrightarrow x(x + 2) = 0$

$\Leftrightarrow \left[\begin{array}{l} x = 0 \quad (\text{loại}) \\ x = -2 \quad (\text{nhận}) \end{array}\right.$

$\Leftrightarrow x = -2$

Vậy $S = \{-2\}$

$c)\ 3x - 8 > 4x - 12$

$\Leftrightarrow 3x - 4x > -12 + 8$

$\Leftrightarrow -x > -4$

$\Leftrightarrow x < 4$

Vậy $S = \{x \mid x < 4\}$

$d)\ \dfrac{x - 1}{2} - \dfrac{7x + 3}{15} \le \dfrac{2x + 1}{3} + \dfrac{3 - 2x}{5}$

$\Leftrightarrow \dfrac{15(x - 1)}{30} - \dfrac{2(7x + 3)}{30} \le \dfrac{10(2x + 1)}{30} + \dfrac{6(3 - 2x)}{30}$

$\Leftrightarrow 15x - 15 - 14x - 6 \le 20x + 10 + 18 - 12x$

$\Leftrightarrow x - 21 \le 8x + 28$

$\Leftrightarrow x - 8x \le 28 + 21$

$\Leftrightarrow -7x \le 49$

$\Leftrightarrow x \ge -7$

Vậy $S = \{x \mid x \ge -7\}$