Giải:
\(\left|-x+\dfrac{2}{5}\right|+\dfrac{1}{2}=3,5\)
\(\Leftrightarrow\left|-x+\dfrac{2}{5}\right|=3,5-\dfrac{1}{2}=3\)
Ta xét các trường hợp:
* Trường hợp 1:
\(-x+\dfrac{2}{5}=3\)
\(\Rightarrow-x=3-\dfrac{2}{5}=2,6\)
\(\Leftrightarrow x=-2,6\)
* Trường hợp 2:
\(-x+\dfrac{2}{5}=-3\)
\(\Rightarrow-x=-3-\dfrac{2}{5}=-3,4\)
\(\Leftrightarrow x=3,4\)
Vậy \(x\in\left\{-2,6;3,4\right\}\)
Chúc bạn học tốt!
$|-x+\dfrac{2}{5}|+\dfrac{1}{2}=3,5$
$=>|-x+0,4|+0,5=3,5$
$=>|-x+0,4|=3,5-0,5=3$
\(=>\left[{}\begin{matrix}-x+0,4=3\\-x+0,4=-3\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}-x=3-0,4=2,6\\-x=-3-0,4=-3,4\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=-2,6\\x=3,4\end{matrix}\right.\)
<=> /-x+2/5/=3.5-0.5=3
<=> -x+2/5=3 => x=-13/5
hoặc -x+2/5=-3=> x=-17/5
vậy x={-13/5;-17/5}
\(\left|-x+\dfrac{2}{5}\right|+\dfrac{1}{2}=3,5\)
\(\left|-x+\dfrac{2}{5}\right|=3,5-\dfrac{1}{2}=3\)
\(\Leftrightarrow-x+\dfrac{2}{5}=3\Rightarrow-x=\dfrac{13}{5}\Leftrightarrow x=-\dfrac{13}{5}\)
\(\Leftrightarrow-x+\dfrac{2}{5}=-3\Rightarrow-x=-\dfrac{17}{5}\Rightarrow x=\dfrac{17}{5}\)
