Đáp án:
a)x ∈ {$\frac{7}{12}$;$\frac{2}{3}$}
b)x ∈ {-$\frac{94}{15}$;$\frac{86}{15}$}
Giải thích các bước giải:
a)$\frac{1}{3}$ -|$\frac{5}{4}$ - 2x| = $\frac{1}{4}$
⇔ |$\frac{5}{4}$ - 2x| = $\frac{1}{12}$
⇔ \(\left[ \begin{array}{l}\frac{5}{4} - 2x=\frac{1}{12}\\\frac{5}{4} - 2x=-\frac{1}{12}\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}2x=\frac{7}{6}\\2x=\frac{4}{3}\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{7}{12}\\x=\frac{2}{3}\end{array} \right.\)
Vậy x ∈ {$\frac{7}{12}$;$\frac{2}{3}$}
b)|x+$\frac{4}{15}$|-|-3,75|=-|-2,15|
⇔ |x+$\frac{4}{15}$|-3,75=-2,15
⇔ |x+$\frac{4}{15}$|=-6
⇔ \(\left[ \begin{array}{l}x+\frac{4}{15}=-6\\x+\frac{4}{15}= 6\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-\frac{94}{15}\\x=\frac{86}{15}\end{array} \right.\)
Vậy x ∈ {-$\frac{94}{15}$;$\frac{86}{15}$}
