a) $\frac{11}{12}$ - ($\frac{2}{5}$ + x ) = $\frac{2}{3}$
⇔ $\frac{2}{5}$ + x = $\frac{1}{4}$
⇔ x = $\frac{1}{4}$ - $\frac{2}{5}$
⇔ x = $\frac{-3}{20}$
Vậy S = { $\frac{-3}{20}$ }
b) 2x . ( x - $\frac{1}{7}$) = 0
Trường hợp 1: 2x = 0 ⇔ x = 0
Trường hợp 2: x - $\frac{1}{7}$ = 0 ⇔ x = $\frac{1}{7}$
Vậy S = { $\frac{1}{7}$; 0 }
d) $\frac{3}{4}$ + $\frac{1}{4}$ : x = 0
⇔ $\frac{1}{4}$ : x = - $\frac{3}{4}$
⇔ x = $\frac{1}{4}$ : - $\frac{3}{4}$
⇔ x = $\frac{-1}{3}$
e) (x - $\frac{3}{4}$) . $\frac{2}{5}$ = -2
⇔ x - $\frac{3}{4}$ = -2 : $\frac{2}{5}$
⇔ x - $\frac{3}{4}$ = -5
⇔ x = -5 + $\frac{3}{4}$
⇔ x = $\frac{-17}{4}$
f) $\frac{2}{3}$x + $\frac{-1}{2}$x + x = $\frac{-5}{36}$
⇔ $\frac{7}{6}$x = $\frac{-5}{36}$
⇔ x = $\frac{-5}{36}$ : $\frac{7}{6}$
⇔ $\frac{-5}{42}$
g) (x + 2)(x - 3) < 0
Trường hợp 1: $\left \{ {{x+2<0} \atop {x-3>0}} \right.$
⇔ $\left \{ {{x<-2} \atop {x>3}} \right.$
⇒ x ∈ ∅
Trường hợp 2: $\left \{ {{x+2>0} \atop {x-3<0}} \right.$
⇔ $\left \{ {{x>-2} \atop {x<3}} \right.$
⇒ -2 < x < 3
Vậy S = { x ∈ Q | -2 < x < 3 }
h) ( x + 1)(x - 3) > 0
Trường hợp 1: $\left \{ {{x+1>0} \atop {x-3<0}} \right.$
⇔ $\left \{ {{x>-1} \atop {x<3}} \right.$
⇒ -1 < x < 3
Trờng hợp 2: $\left \{ {{x+1<0} \atop {x-3>0}} \right.$
⇔ $\left \{ {{x<-1} \atop {x>3}} \right.$
⇒ x ∈ ∅
Vậy S = { x ∈ Q | -1 < x < 3 }
