$\begin{array}{l} \Leftrightarrow \frac{3. (2 x + 3)}{12} - \frac{2. (5 x + 3)}{12} = \frac{3 - 4 x}{12} \\ \Leftrightarrow \frac{6 x + 9 - 10 x - 6}{12} = \frac{3 - 4 x}{12} \\ \Leftrightarrow \frac{3 - 4 x}{12} = \frac{3 - 4 x}{12} \\ \Leftrightarrow 3 - 4 x = 3 - 4 x \\ \Leftrightarrow - 4 x + 4 x = 3 - 3 \\ \Leftrightarrow 0. x = 0 \\ \Rightarrow x \in ℝ \\ S = \{x | x \in ℝ\} \end{array}$

Câu 3/30

3. $(2 x + 5) (x - 4) + 3 (x - 4) = 0$

Lời giải

$(2 x + 5) (x - 4) + 3 (x - 4) = 0$

$\begin{array}{l} \Leftrightarrow (x - 4) (2 x + 5 + 3) = 0 \\ \Leftrightarrow (x - 4) (2 x + 8) = 0 \\ \Leftrightarrow [\begin{array}{l} x - 4 = 0 \\ 2 x + 8 = 0 \end{array} \Leftrightarrow [\begin{array}{l} x = 4 \\ x = - 4 \end{array} \\ S = \{- 4; 4\} \end{array}$

Câu 4/30

4. $2 {(3 x + 5)}^{2} = (3 x + 5) (x - 7)$

Lời giải

$2 {(3 x + 5)}^{2} = (3 x + 5) (x - 7)$

$\begin{array}{l} \Leftrightarrow (3 x + 5) (6 x + 10 - x + 7) = 0 \\ \Leftrightarrow (3 x + 5) (5 x + 17) = 0 \\ \Leftrightarrow [\begin{array}{l} 3 x + 5 = 0 \\ 5 x + 17 = 0 \end{array} \Leftrightarrow [\begin{array}{l} x = \frac{- 5}{3} \\ x = \frac{- 17}{5} \end{array} \\ S = \{\frac{- 5}{3}; \frac{- 17}{5}\} \end{array}$

Câu 5/30

$5 . 9 {(2 x - 1)}^{2} - 4 {(x + 1)}^{2} = 0$

Lời giải

$9 {(2 x - 1)}^{2} - 4 {(x + 1)}^{2} = 0$

$\begin{array}{l} \Leftrightarrow 9 (4 x^{2} - 4 x + 1) - 4 (x^{2} + 2 x + 1) = 0 \\ \Leftrightarrow {[3 (2 x - 1)]}^{2} - {[2 (x + 1)]}^{2} = 0 \\ \Leftrightarrow (6 x - 3 + 2 x + 2) (6 x - 3 - 2 x - 2) = 0 \\ \Leftrightarrow (4 x - 5) (8 x - 1) = 0 \\ \Leftrightarrow [\begin{array}{l} 4 x - 5 = 0 \\ 8 x - 1 = 0 \end{array} \Leftrightarrow [\begin{array}{l} x = \frac{5}{4} \\ x = \frac{1}{8} \end{array} \end{array}$

$S = \{\frac{5}{4}; \frac{1}{8}\}$

Câu 6/30

6. $x^{3} - 2 x^{2} + 2 - x = 0$

Lời giải

$x^{3} - 2 x^{2} + 2 - x = 0$

$\begin{array}{l} \Leftrightarrow x^{2} (x - 2) - (x - 2) = 0 \\ \Leftrightarrow (x - 2) (x^{2} - 1) = 0 \\ \Leftrightarrow (x + 1) (x - 2) (x - 1) = 0 \\ \Leftrightarrow [\begin{array}{l} x + 1 = 0 \\ x - 2 = 0 \\ x - 1 = 0 \end{array} \Leftrightarrow [\begin{array}{l} x = - 1 \\ x = 2 \\ x = 1 \end{array} \end{array}$

Vậy, $S = \{- 1; 1; 2\}$

Câu 7/30

7. $x^{2} - x - 12 = 0$

Lời giải

$x^{2} - x - 12 = 0$

$\begin{array}{l} \Leftrightarrow x^{2} - 4 x + 3 x - 12 = 0 \\ \Leftrightarrow x (x - 4) + 3 (x - 4) = 0 \\ \Leftrightarrow (x - 4) (x + 3) = 0 \\ \Leftrightarrow [\begin{array}{l} x - 4 = 0 \\ x + 3 = 0 \end{array} \Leftrightarrow [\begin{array}{l} x = 4 \\ x = - 3 \end{array} \end{array}$

$S = \{- 3; 4\}$

Câu 8/30

8 . $\frac{1}{x - 1} + \frac{2}{x + 1} = \frac{x}{x^{2} - 1}$

Lời giải

$\frac{1}{x - 1} + \frac{2}{x + 1} = \frac{x}{x^{2} - 1}$ (ĐKXĐ: $x \neq \pm 1$ )

$\begin{array}{l} \Leftrightarrow \frac{x + 1 + 2 (x - 1)}{(x - 1) (x + 1)} = \frac{x}{x^{2} - 1} \\ \Leftrightarrow \frac{3 x - 1}{(x - 1) (x + 1)} = \frac{x}{(x - 1) (x + 1)} \\ \Rightarrow 3 x - 1 = x \\ \Leftrightarrow 2 x = 1 \\ \Leftrightarrow x = \frac{1}{2} (T M) \end{array}$