$a)$
$|x^{2}-1|=12+3^5:3^4$
$⇔|x^{2}-1|=12+3^{(5-4)}$
$⇔|x^{2}-1|=12+3^{1}$
$⇔|x^{2}-1|=15$
\(⇔\left[ \begin{array}{l}x^2-1=15\\x^2-1=-15\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x^2=16\\x^2=-14(vô lí vì x^2≥0 ∀x)\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=4\\x=-4\end{array} \right.\)
Vậy..
$b)$
$(x-\frac{1}{2})^{2}=\frac{4}{25}$
\(⇔\left[ \begin{array}{l}x-\frac{1}{2}=\frac{2}{5}\\x-\frac{1}{2}=-\frac{2}{5}\end{array}
\right.\)
\(⇔\left[ \begin{array}{l}x=\frac{9}{10}\\x=\frac{1}{10}\end{array} \right.\)
Vậy
