Câu 1:
a) 2x - 6 = 0
⇔ 2x = 6
⇔ x = 3
Vậy S = {3}
b) (x + 2)(3x - 10) = 0
⇔ x + 2 = 0 hay 3x - 10 = 0
⇔ x = -2 hay 3x = 10
⇔ x = -2 hay x = $\frac{10}{3}$
Vậy S = {$\frac{10}{3}$}
c) $\frac{x+5}{x-5}$ - $\frac{x-5}{x+5}$ = $\frac{20}{x²-25}$
⇔ $\frac{x+5}{x-5}$ - $\frac{x-5}{x+5}$ = $\frac{20}{(x-5)(x+5)}$
Điều kiện: x - 5 ∦ 0 ⇔ x ∦ 5
x + 5 ∦ 0 ⇔ x ∦ -5
$\frac{(x+5)²}{(x-5)(x+5)}$ - $\frac{(x-5)²}{(x-5)(x+5)}$ = $\frac{20}{(x-5)(x+5)}$
⇔ x² + 10x + 25 - x² + 10x - 25 = 20
⇔ 20x = 20
⇔ x = 1
Vậy S = {1}
Câu 2:
a) x - 2 > 0
⇔ x > 2
Vậy S = {x/x>2}
b) 6x - 2 < 4(x+1)
⇔ 6x - 2 < 4x + 4
⇔ 2x < 6
⇔ x < 3
Vậy S = (x/x < 3)
c) $\frac{2-x}{2}$ > $\frac{3x-1}{3}$
⇔ $\frac{6-3x}{6}$ > $\frac{6x-2}{6}$
⇔ 6 - 3x > 6x - 2
⇔ -9x > -8
⇔ x > $\frac{8}{9}$
