Đáp án:
Giải thích các bước giải:
$b)\frac{3}{2}-|1-\frac{1}{4}+3x|=\frac{1}{4}_{}$
⇔ $\frac{3}{2}-|\frac{3}{4}+3x|=\frac{1}{4}_{}$
⇔ $-|\frac{3}{4}+3x|=\frac{1}{4}-\frac{3}{2}_{}$
⇔ $-|\frac{3}{4}+3x=-\frac{5}{4}_{}$
⇔ $|\frac{3}{4}+3x|=\frac{5}{4}_{}$
⇔ \(\left[ \begin{array}{l}\frac{3}{4}+3x=\frac{5}{4}\\\frac{3}{4}+3x=-\frac{5}{4}\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=\frac{1}{6}\\x=-\frac{2}{3}\end{array} \right.\)
$Vậy _{}$ $x=_{}$ $\frac{1}{6}$ $và_{}$ $-\frac{2}{3}$
$d)|x-1|-2x=\frac{1}{2}_{}$
⇔ \(\left[ \begin{array}{l}x-1-2x=\frac{1}{2},x-1\geq0\\-(x-1)-2x=\frac{1}{2},x-1<0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=-\frac{3}{2},x\geq1\\x=\frac{1}{6},x<1\end{array} \right.\)
$Vậy_{}$ $x=_{}$ $\frac{1}{6}$
