Bài 2. (1,5 điểm) Giải các phương trình sau: 1) $2(4-3x)=5x-14$ 2) $\frac{x+3}{x-4}$ +$\frac{2-x^{2}}{(x-4)(x-2)}$ =$\frac{x-1}{2-x}$ 

$\text{1)}$ `2(4 - 3x) = 5x - 14`

$\Leftrightarrow$ `8 - 6x = 5x - 14`

$\Leftrightarrow$ `-6x - 5x = -14 - 8`

$\Leftrightarrow$ `-11x = -22`

$\Leftrightarrow$ `x = 2`

$\text{Vậy S =}$ `{2}`

$\text{2)}$ `\frac{x + 3}{x - 4} + \frac{2 - x²}{(x - 4)(x - 2)} = \frac{x - 1}{2 - x}`

$\text{(đkxđ:}$ `x \ne 2; x \ne 4)`

$\Leftrightarrow$ `\frac{(x - 2)(x + 3)}{(x - 2)(x - 4)} + \frac{2 - x²}{(x - 4)(x - 2)} = -\frac{x - 1}{x - 2}`

$\Leftrightarrow$ `\frac{(x - 2)(x + 3)}{(x - 2)(x - 4)} + \frac{2 - x²}{(x - 4)(x - 2)} = -\frac{(x - 4)(x - 1)}{(x - 4)(x - 2)}`

$\Leftrightarrow$ `(x - 2)(x + 3) + 2 - x² = -(x - 4)(x - 1)`

$\Leftrightarrow$ `x² + x - 6 + 2 - x²= -x² + 5x - 4`

$\Leftrightarrow$ `x² - 4x = 0`

$\Leftrightarrow$ `x(x - 4) = 0`

$\Leftrightarrow$ $\left[\begin{matrix} x = 0 (TM)\\ x - 4 = 0\end{matrix}\right.$

$\Leftrightarrow$ $\left[\begin{matrix} x = 0 (TM)\\ x = 4(KTM)\end{matrix}\right.$

$\text{Vậy S =}$ `{0}`