$b$) $|2x+3|=4x+1$ ($1$)
$TH1:$ $2x+3≥0⇔2x≥-3⇔x≥-3/2$
Từ ($1$),ta có: $2x+3=4x+1$
$⇔ 3-1=4x-2x$
$⇔ 2 = 2x$
$⇔ x =1$ ($\Rightarrow TM$)
$TH2:$ $2x+3<0⇔2x<-3⇔x<-3/2$
Từ ($1$),ta có: $-(2x+3)=4x+1$
$⇔ -2x-3=4x+1$
$⇔ -3 - 1 = 4x - (-2x)$
$⇔ -4 = 6x$
$⇔ x = -2/3($\Rightarrow KTM$)
Vậy $x=1$