a, \(\left|x+1\right|-4=0\)
\(\Rightarrow\left|x+1\right|=4\)
+) Xét \(x\ge-1\) có:
\(x+1=4\Rightarrow x=3\) ( t/m )
+) Xét x < -1 có:
\(-x-1=4\Rightarrow x=-5\) ( t/m )
Vậy x = 3 hoặc x = -5
b, tương tự
c, \(3x+\left|2x\right|=5x\)
\(\Rightarrow\left|2x\right|=2x\)
+) Xét \(x\ge0\)
\(\Rightarrow2x=2x\Rightarrow x\in R\forall x>0\)
+) Xét x < 0
\(\Rightarrow-2x=2x\Rightarrow x=0\)
Vậy \(x\ge0\)
d, \(\left|2x+1\right|=\left|x-3\right|\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=x-3\\2x+1=3-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy...