Đáp án:
$\text{ c) | x+ $\dfrac{4}{5}$ | - $\dfrac{1}{7}$ = 0 }$
$\text{TH1 : }$
$\text{⇒x+$\dfrac{4}{5}$ ≥ 0 ⇔ x ≥ $\dfrac{-4}{5}$ }$
$\text{⇔( x + $\dfrac{4}{5}$) - $\dfrac{1}{7}$ = 0 }$
$\text{⇔ x +$\dfrac{4}{5}$ = $\dfrac{1}{7}$ }$
$\text{⇔ x = $\dfrac{1}{7}$ - $\dfrac{4}{5}$ }$
$\text{⇔ x = -$\dfrac{23}{35}$ (Thỏa mãn) }$
$\text{Th2}$
$\text{⇒ x + $\dfrac{4}{5}$ < 0 ⇔ x < -$\dfrac{4}{5}$ }$
$\text{⇔ - (x+$\dfrac{4}{5}$) - $\dfrac{1}{7}$ = 0}$
$\text{⇔ - (x + $\dfrac{4}{5}$) = $\dfrac{1}{7}$ }$
$\text{⇔ -x - $\dfrac{4}{5}$ = $\dfrac{1}{7}$ }$
$\text{⇔ -x = $\dfrac{1}{7}$ + $\dfrac{4}{5}$ }$
$\text{⇔ -x=$\dfrac{33}{35}$ }$
$\text{⇔ x = -$\dfrac{33}{35}$ (Thỏa mãn) }$
$\text{Vậy x ∈ { $\dfrac{23}{35}$ ; -$\dfrac{33}{35}$ } }$
$\text{d) (2x+1)³ - 8 = 0 }$
$\text{⇔ (2x+1)³ - 2³ =0 }$
$\text{⇔ (2x+1) - 2 = 0 }$
$\text{⇔ 2x + 1 = 2 }$
$\text{⇔ 2x = 2 - 1 }$
$\text{⇔ 2x = 1}$
$\text{⇔ x = $\dfrac{1}{2}$ }$
$\text{Vậy x =$\dfrac{1}{2}$ }$
