Giúp em vs ạ 9h em phải nộp rôifl

Đáp án:

a) $x=\pm6$

b) $x=\dfrac{1}{2}$

c) $x=1$ hoặc $x=-\dfrac{3}{2}$

Giải thích các bước giải:

a)

$|x|+6=12\\⇔|x|=6\\⇔x=\pm6$

Vậy $x=\pm6$

b)

$\left|\dfrac{1}{2}x-\dfrac{1}{4}\right|=0\\⇔\dfrac{1}{2}x-\dfrac{1}{4}=0\\⇔\dfrac{1}{2}x=\dfrac{1}{4}\\⇔4x=2\\⇔x=\dfrac{1}{2}$

Vậy $x=\dfrac{1}{2}$

c)

$\left|2x+\dfrac{1}{2}\right|=\dfrac{5}{2}\\⇔\left[\begin{array}{l}2x+\dfrac{1}{2}=\dfrac{5}{2}\\2x+\dfrac{1}{2}=-\dfrac{5}{2}\end{array}\right.\\⇔\left[\begin{array}{l}2x=2\\2x=-3\end{array}\right.\\⇔\left[\begin{array}{l}x=1\\x=-\dfrac{3}{2}\end{array}\right.$

Vậy $x=1$ hoặc $x=-\dfrac{3}{2}$