Đáp án:
a) $x∈\left\{\dfrac{7}{12};\dfrac{13}{12}\right\}$
b) $x=±\dfrac{23}{28}$
Giải thích các bước giải:
$a)\left|\dfrac{5}{6}-x\right|=\dfrac{1}{4}$
Trường hợp 1:
$\dfrac{5}{6}-x=\dfrac{1}{4}$ $\left(x≤\dfrac{5}{6}\right)$
$⇒-x=\dfrac{1}{4}-\dfrac{5}{6}$
$⇒-x=-\dfrac{7}{12}$
$⇒x=\dfrac{7}{12}(tm)$
Trường hợp 2:
$x-\dfrac{5}{6}=\dfrac{1}{4}$ $\left(x>\dfrac{5}{6}\right)$
$⇒x=\dfrac{5}{6}+\dfrac{1}{4}$
$⇒x=\dfrac{13}{12}(tm)$
Vậy $x∈\left\{\dfrac{7}{12};\dfrac{13}{12}\right\}$
$b)|x|-\dfrac{1}{4}=\dfrac{4}{7}$
$⇒|x|=\dfrac{1}{4}+\dfrac{4}{7}$
$⇒|x|=\dfrac{23}{28}$
$⇒x=±\dfrac{23}{28}$
Vậy $x=±\dfrac{23}{28}$
