a: A=5x+3+|3-x|
=5x+3+|x-3|
TH1: x>=3
=>A=5x+3+x-3=6x
TH2: x<3
=>A=5x+3+3-x=6+4x
b: B=2x-5-|2x+1|
TH1: \(x>=-\dfrac{1}{2}\)
=>B=2x-5-(2x+1)=2x-5-2x-1=-6
TH2: \(x< -\dfrac{1}{2}\)
=>B=2x-5-(-2x-1)=2x-5+2x+1=4x-4
c: C=|3-2x|+2x-3
=|2x-3|+2x-3
TH1: \(x>=\dfrac{3}{2}\)
=>C=2x-3+2x-3=4x-6
TH2: \(x< \dfrac{3}{2}\)
=>|2x-3|=3-2x
=>C=3-2x+2x-3=0
d: Đặt A=|4x+5|-4x+2
TH1: \(x>=-\dfrac{5}{4}\)
=>|4x+5|=4x+5
A=|4x+5|-4x+2=4x+5-4x+2=7
TH2: \(x< -\dfrac{5}{4}\)
=>|4x+5|=-4x-5
A=|4x+5|-4x+2=-4x-5-4x+2=-8x-3