1c) $\frac{x-5}{x-1}$ + $\frac{2}{x-3}$ = 1 ĐK: x $\neq$ 1;3
⇔ $\frac{(x-5)(x-3)}{(x-1)(x-3)}$ + $\frac{2(x-1)}{(x-3)(x-1)}$ = $\frac{(x-1)(x-3)}{(x-1)(x-3)}$
⇔ (x-5)(x-3) + 2(x-1) = (x-1)(x-3)
⇔ x² -3x -5x +15 +2x -2 = x² -3x -x +3
⇔ x² - x² -3x -5x +2x +3x +x = 3 -15 +2
⇔ -2x = -10
⇔ x = 5 (nhận)
S = {5}
2b) $\frac{2-4x}{5}$ < $\frac{-x+4}{3}$
⇔ $\frac{3(2-4x)}{5.3}$ < $\frac{5(-x+4)}{3.5}$
⇔ 3(2-4x) < 5(-x+4)
⇔ 6 -12x < -5x +20
⇔ -12x +5x < 20 -6
⇔ -7x < 14
⇒ x < -2
S={ x | x < -2 }
