`2)`
`a)`
`\sqrt{(2x+1)^2}=3`
`<=>|2x+1|=3`
`<=>`\(\left[ \begin{array}{l}2x+1=3\\2x+1=-3\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=1\\x=-2\end{array} \right.\)
Vậy `S={1;-2}`
`b)`
`\sqrt{(3x-1)^2}=2`
`<=>|3x-1|=2`
`<=>`\(\left[ \begin{array}{l}3x-1=2\\3x-1=-2\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=1\\x=-\dfrac{1}{3}\end{array} \right.\)
Vậy `S={1;-1/3}`
`c)`
`\sqrt{x^2+2x+1}=5`
`<=>\sqrt{(x+1)^2}=5`
`<=>|x+1|=5`
`<=>`\(\left[ \begin{array}{l}x+1=5\\x+1=-5\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=4\\x=-6\end{array} \right.\)
Vậy `S={4;-6}`
`d)`
`\sqrt{x^2-4x+4}=1`
`<=>\sqrt{(x-2)^2}=1`
`<=>|x-2|=1`
`<=>`\(\left[ \begin{array}{l}x-2=1\\x-2=-1\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=3\\x=1\end{array} \right.\)
Vậy `S={3;1}`
