1: |2x-1|=|x+3|
=>\(\left[\begin{array}{l}2x-1=x+3\\ 2x-1=-x-3\end{array}\right.\Rightarrow\left[\begin{array}{l}2x-x=3+1\\ 2x+x=-3+1\end{array}\right.\)
=>\(\left[\begin{array}{l}x=4\\ 3x=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=-\frac23\end{array}\right.\)
2: \(\left|x^2-7x+12\right|=\left|x-4\right|\)
=>\(\left[\begin{array}{l}x^2-7x+12=x-4\\ x^2-7x+12=-\left(x-4\right)\end{array}\right.\Rightarrow\left[\begin{array}{l}\left(x-4\right)\left(x-3\right)-\left(x-4\right)=0\\ \left(x-3\right)\left(x-4\right)+\left(x-4\right)=0\end{array}\right.\)
=>\(\left[\begin{array}{l}\left(x-4\right)^2=0\\ \left(x-4\right)\left(x-2\right)=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=2\end{array}\right.\)
3:|2-x|=2x-1
=>\(\left|x-2\right|=2x-1\)
=>\(\begin{cases}\left(2x-1\right)^2=\left(x-2\right)^2\\ 2x-1\ge0\end{cases}\Rightarrow\begin{cases}\left(2x-1-x+2\right)\left(2x-1+x-2\right)=0\\ 2x\ge1\end{cases}\)
=>\(\begin{cases}\left(x+1\right)\left(3x-3\right)=0\\ x\ge\frac12\end{cases}\Rightarrow x=1\)
4:\(\left|2x+5\right|-x=2\)
=>|2x+5|=x+2
=>\(\begin{cases}x+2\ge0\\ \left(2x+5\right)^2=\left(x+2\right)^2\end{cases}\Rightarrow\begin{cases}x\ge-2\\ \left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\end{cases}\)
=>\(\begin{cases}x\ge-2\\ \left(x+3\right)\left(3x+7\right)=0\end{cases}\Rightarrow x\in\) ∅
6: \(\left|x^2-x+5\right|=2x-1\)
=>\(x^2-x+5=2x-1\)
=>\(x^2-3x+6=0\)
\(\Delta=\left(-3\right)^2-4\cdot1\cdot6=9-24=-15<0\)
=>Phương trình vô nghiệm
