Giải:
\(\dfrac{17}{2}-\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)
\(\Leftrightarrow\left|2x-\dfrac{3}{4}\right|=\dfrac{17}{2}-\left(-\dfrac{7}{4}\right)\)
\(\Leftrightarrow\left|2x-\dfrac{3}{4}\right|=\dfrac{17}{2}+\dfrac{7}{4}\)
\(\Leftrightarrow\left|2x-\dfrac{3}{4}\right|=\dfrac{41}{4}\)
Xét các trường hợp:
* \(2x-\dfrac{3}{4}=\dfrac{41}{4}\)
\(\Leftrightarrow2x=\dfrac{41}{4}+\dfrac{3}{4}\)
\(\Leftrightarrow2x=11\)
\(\Leftrightarrow x=5,5\)
* \(2x-\dfrac{3}{4}=-\dfrac{41}{4}\)
\(\Leftrightarrow2x=-\dfrac{41}{4}+\dfrac{3}{4}\)
\(\Leftrightarrow2x=-\dfrac{19}{2}\)
\(\Leftrightarrow x=-\dfrac{19}{4}\)
Vậy ...
\(\dfrac{17}{2}-\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\\ \left|2x-\dfrac{3}{4}\right|=\dfrac{41}{4}\\ \left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{41}{4}\\2x-\dfrac{3}{4}=-\dfrac{41}{4}\end{matrix}\right.\\ \left[{}\begin{matrix}x=\dfrac{11}{2}\\x=-\dfrac{19}{4}\end{matrix}\right.\)
/2x-\(\dfrac{3}{4}\)/=\(\dfrac{17}{2}-\dfrac{-7}{4}\)
2x-\(\dfrac{3}{4}\)=\(\dfrac{41}{4}\)
2x=\(\dfrac{41}{4}\)+\(\dfrac{3}{4}\)
2x=11
x=11:2
x=\(\dfrac{11}{2}\)
