a) `x/(x - 3) - (x - 1)/x = 1`

⇔ `x.(x - 1)/(x - 3)(x - 1) - (x - 1)(x - 3)/x.(x - 3) = (x - 1)(x - 3)/(x - 1)(x - 3)`

⇒ `x.(x - 1) - (x - 1)(x - 3) = (x - 1)(x - 3)`

⇔ `x² - x - x² + 4x - 3 = x² - 4x + 3`

⇔ `x² - x - x² + 4x - 3 - x² + 4x - 3 = 0`

⇔ `-x² + 7x - 6 = 0`

⇔ `x² - 7x + 6 = 0`

⇔ `x² - x - 6x + 6 = 0`

⇔ `(x² - x) - (6x - 6) = 0`

⇔ `x. (x - 6) - 6. (x - 1) = 0`

⇔ `(x - 6) (x - 1) = 0`

⇔ $\left[\begin{matrix} x - 6 = 0\\ x - 1 = 0\end{matrix}\right.$

⇔ $\left[\begin{matrix} x = 6\\ x = 1\end{matrix}\right.$

Vậy `S = {6 ; 1}`

b) `(x - 6) (x - 3) = 2(x - 3)`

⇔ `(x - 6) (x - 3) - 2(x - 3) = 0`

⇔ `(x - 6 - 2) (x - 3) = 0`

⇔ `(x - 8) (x - 3) = 0`

⇔ $\left[\begin{matrix} x - 8 = 0\\ x - 3 = 0\end{matrix}\right.$

⇔ $\left[\begin{matrix} x = 8\\ x = 3\end{matrix}\right.$

Vậy `S = {8 ; 3}`

c) `(2x + 3)/-4 ≥ 4 - x/4`

⇔ `(-2x - 3)/4 ≥ 4/4 - x/4`

⇒ `-2x - 3 ≥ 4 - x`

⇔ `-2x + x ≥ 3 + 4`

⇔ `-x ≥ 7`

⇔ `x < -7`

Vậy `x < -7`

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