Bạn tham khảo:
a. `\frac{1}{2}-(\frac{2}{3}x-\frac{1}{3})=\frac{2}{3}`.
$⇔\dfrac{1}{2}-\dfrac{2x}{3}+\dfrac{1}{3}=\dfrac{2}{3}$.
$⇔-\dfrac{2x}{3}+\dfrac{5}{6}=\dfrac{2}{3}$.
$⇔-\dfrac{2x}{3}=-\dfrac{1}{6}$.
$⇔-2x=-\dfrac{1}{2}$.
$⇔x=\dfrac{1}{4}$.
Vậy: $x=\dfrac{1}{4}$.
b. $\dfrac{3}{x+5}=15\%$. (Điều kiện: $x \neq -5$).
$⇔\dfrac{3}{x+5}=\dfrac{3}{20}$.
$⇔\dfrac{60}{20(x+5)}=\dfrac{3(x+5)}{20(x+5)}$.
$⇔60=3x+15$.
$⇔-3x=-45$.
$⇔x=15$.
Vậy: $x=15$.
