Bài 1:
a: Khi x>=0 thì 5x>=0
=>|5x|=5x
\(A=3x+2+\left|5x\right|\)
=3x+2+5x
=8x+2
b: Khi x>0 thì -4x<0
=>|-4x|=4x
\(B=\left|-4x\right|-2x+12\)
=4x-2x+12
=2x+12
c: Khi x>5 thì x-4>5-4=1>0
=>|x-4|=x-4
\(C=\left|x-4\right|-2x+12=x-4-2x+12=-x+8\)
Bài 2:
a: \(\left|2x\right|=x-6\)
=>\(\left\{{}\begin{matrix}x-6>=0\\\left(2x\right)^2=\left(x-6\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=6\\\left(2x-x+6\right)\left(2x+x-6\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=6\\\left(x+6\right)\left(3x-6\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
b: \(\left|-3x\right|=x-8\)
=>\(\left|3x\right|=x-8\)
=>\(\left\{{}\begin{matrix}x>=8\\\left(3x\right)^2=\left(x-8\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=8\\\left(3x-x+8\right)\left(3x+x-8\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=8\\\left(2x+8\right)\left(4x-8\right)=0\end{matrix}\right.\)
=>\(x\in\varnothing\)
c: |x+3|=3x-1
=>\(\left\{{}\begin{matrix}3x-1>=0\\\left(3x-1\right)^2=\left(x+3\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{1}{3}\\\left(3x-1-x-3\right)\left(3x-1+x+3\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{1}{3}\\\left(2x-4\right)\left(4x+2\right)=0\end{matrix}\right.\)
=>x=2
d: |x-4|=-3x+5
=>\(\left\{{}\begin{matrix}-3x+5>=0\\\left(-3x+5\right)^2=\left(x-4\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{5}{3}\\\left(3x-5-x+4\right)\left(3x-5+x-4\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{5}{3}\\\left(2x-1\right)\left(4x-9\right)=0\end{matrix}\right.\)
=>\(x=\dfrac{1}{2}\)
