\(TH1:\left|x+1\right|=x+1\\ \Leftrightarrow2x-\left(x+1\right)=-\dfrac{1}{2}\\ \Leftrightarrow2x-x-1=-\dfrac{1}{2}\\ \Leftrightarrow x-1=-\dfrac{1}{2}\\ \Leftrightarrow x=-\dfrac{1}{2}+1\\ \Leftrightarrow x=\dfrac{1}{2}\\ TH2:\left|x+1\right|=-x-1\\ \Leftrightarrow2x-\left(-x-1\right)=-\dfrac{1}{2}\\ \Leftrightarrow2x+x+1=-\dfrac{1}{2}\\ \Leftrightarrow3x=-\dfrac{1}{2}-1\\ \Leftrightarrow3x=-\dfrac{3}{2}\\ \Leftrightarrow x=\left(-\dfrac{3}{2}\right):3=-\dfrac{1}{2}\)
Thay lần lượt \(x=\dfrac{1}{2};x=-\dfrac{1}{2}\) vào pt
\(\Rightarrow x=\dfrac{1}{2}\left(thoaman\right)\)
Vậy \(x=\dfrac{1}{2}\)
\(2x-\left|x+1\right|=-\dfrac{1}{2}\)
\(\left|x+1\right|=2x+\dfrac{1}{2}\)
\(\left[{}\begin{matrix}x+1=2x+\dfrac{1}{2}\\x+1=-2x-\dfrac{1}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{2};-\dfrac{1}{2}\right\}\).
2x - | x + 1| = \(-\dfrac{1}{2}\) ⇒ 2x + \(\dfrac{1}{2}\) = | x + 1| đk 2x + \(\dfrac{1}{2}\) ≥ 0 ⇒ x ≥ - \(\dfrac{1}{4}\)
Với x ≥ \(-\dfrac{1}{4}\) ta có : 2x + \(\dfrac{1}{2}\) = x + 1 ⇒ 2x - x = 1 - \(\dfrac{1}{2}\) ⇒ x = \(\dfrac{1}{2}\) (thỏa mãn)
vậy x = \(\dfrac{1}{2}\)
