${g,|3x-}$ $\dfrac{1}{2}|+$ $\dfrac{1}{4}=$ $\dfrac{5}{4}$
${|3x-}$ $\dfrac{1}{2}|=$ $\dfrac{5}{4}-$ $\dfrac{1}{4}$
${|3x-}$ $\dfrac{1}{2}|=1$
${=>}$ \(\left[ \begin{array}{l}3x-1/2=1\\3x-1/2=-1\end{array} \right.\)
${=>}$ \(\left[ \begin{array}{l}x=1/2\\x=-1/6\end{array} \right.\)
Vậy ${x}$ ∈ {$\dfrac{1}{2};$ $\dfrac{-1}{6}$}
${h,}$ $\dfrac{-4}{|-3|}=$ $\dfrac{-12}{|x-2|}$
${=>-4.|x-2|=-12.|-3|}$
${=>-4.|x-2|=-12.3}$
${=>-4.|x-2|=-36}$
${|x-2|=-36:(-4)}$
${|x-2|=9}$
${=>}$ \(\left[ \begin{array}{l}x-2=9\\x-2=-9\end{array} \right.\)
${=>}$ \(\left[ \begin{array}{l}x=11\\x=-7\end{array} \right.\)
Vậy ${x}$ ∈ {${11;-7}$}
~Xin hay nhất~
