Trả lời bài 2 hộ em với ạ

Bài 2 :

$a.\dfrac{-3}{2}x+\dfrac{3}{4}=\dfrac{7}{8} \\⇔\dfrac{-3}{2}x=\dfrac{7}{8}-\dfrac{3}{4} \\⇔\dfrac{-3}{2}x=\dfrac{1}{8} \\⇔x=\dfrac{1}{8}:\dfrac{-3}{2} \\⇔x=\dfrac{-1}{12}$

$Vậy\ x=\dfrac{-1}{12}$

$b.\bigg(2x-\dfrac{1}{2}\bigg)^2=\dfrac{1}{16}$

$⇔2x-\dfrac{1}{2}=±\dfrac{1}{4}$

⇔\(\left[ \begin{array}{l}2x-\dfrac{1}{2}=\dfrac{1}{4}\\2x-\dfrac{1}{2}=\dfrac{-1}{4}\end{array} \right.\) ⇔\(\left[ \begin{array}{l}2x=\dfrac{3}{4}\\2x=\dfrac{1}{4}\end{array} \right.\)  ⇔\(\left[ \begin{array}{l}x=\dfrac{3}{8}\\x=\dfrac{1}{8}\end{array} \right.\)

Vậy \(\left[ \begin{array}{l}x=\dfrac{3}{8}\\x=\dfrac{1}{8}\end{array} \right.\)

$c.(2x+3)\bigg(\dfrac{1}{2}x-\dfrac{3}{2}\bigg)=0$

⇔\(\left[ \begin{array}{l}2x+3=0\\\dfrac{1}{2}x-\dfrac{3}{2}=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}2x=-3\\\dfrac{1}{2}x=\dfrac{3}{2}\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=\dfrac{-3}{2}\\x=3\end{array} \right.\)

Vậy \(\left[ \begin{array}{l}x=\dfrac{-3}{2}\\x=3\end{array} \right.\)