\(\left|2x-1\right|=x+4\)
+) Xét \(x\ge\frac{1}{2}\) có:
\(2x-1=x+4\)
\(\Rightarrow x=5\) ( t/m )
+) Xét \(x< \frac{1}{2}\) có:
\(1-2x=x+4\)
\(\Rightarrow-x=-3\)
\(\Rightarrow x=3\) ( loại )
Vậy x = 5
Đk: \(x+4\ge0\Rightarrow x\ge-4\)
\(\left|2x+1\right|=x+4\)
\(\Rightarrow\left[\begin{matrix}2x-1=x+4\\2x-1=-x-4\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}2x-x=4+1\\2x+x=-4+1\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=5\left(tm\right)\\x=-1\left(loai\right)\end{matrix}\right.\)
Vậy x=5
Vì \(\left|2x-1\right|\ge0\) Để \(\left|2x-1\right|=x+4\) => \(x+4\ge0\Rightarrow x\ge-4\)
|2x - 1| = x + 4 => 2x - 1 = \(\pm\left(x+4\right)\)
TH1 : 2x - 1 = x + 4 <=> 2x - x = 4 + 1 => x = 5 (TM)
TH2 : 2x - 1 = - x - 4 <=> 2x + x = - 4 + 1 <=> 3x = - 3 => x = - 1(TM)
Vậy x = { - 1; 5 }