Câu trả lời:

\(\sqrt{\left(2x+1\right)^2}\)=3( ĐKXĐ: 2x+1\(\ge\)0<=>x\(\ge\)-\(\dfrac{1}{2}\)

/2x+1/=3

TH1: 2x+1=3

<=>2x= 4

<=> x=2(TM)

TH2: 2x+1=-3

<=> 2x= -2

<=> x=-1(TM)