d) $\frac{5}{3}$ -$|x-\frac{1}{4}|=$ $\frac{1}{6}$+ $\frac{2}{3}$
$\frac{5}{3}$ -$|x-\frac{1}{4}|=$ $\frac{5}{6}$
$|x-\frac{1}{4}|=$ $\frac{5}{3}$ - $\frac{5}{6}$
$|x-\frac{1}{4}|=$ $\frac{5}{6}$
⇒\(\left[ \begin{array}{l}x-\frac{1}{4}=\frac{5}{6}\\x-\frac{1}{4}=\frac{-5}{6}\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=\frac{1}{4}+\frac{5}{6}\\x=\frac{1}{4}-\frac{5}{6}\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=\frac{13}{12}\\x=\frac{-7}{12}\end{array} \right.\)
c) $|x-\frac{1}{4}|$+$1\frac{1}{3}=$ $\frac{11}{6}$- |$\frac{-3}{4}|$
$|x-\frac{1}{4}|$=$\frac{11}{6}$-$\frac{3}{4}$-$1\frac{1}{3}$
$|x-\frac{1}{4}|$=$\frac{11}{6}$-$\frac{3}{4}$-$\frac{4}{3}$
$|x-\frac{1}{4}|$=$\frac{-1}{4}$
Do $|x-\frac{1}{4}|$≥0 với mọi x
mà $|x-\frac{1}{4}|$=$\frac{-1}{4}$
=> x không có giá trị nào thỏa mãn
