Ta có :

a.  $\sqrt{(x-3)^{2}}=9$

<=> $\left | x-3 \right |=9$

<=> $\left\{\begin{matrix}x-3 =9& \\ x-3=-9 & \end{matrix}\right.<=> \left\{\begin{matrix}x=12 & \\ x=-6 & \end{matrix}\right.$

Vậy $ \left\{\begin{matrix}x=12 & \\ x=-6 & \end{matrix}\right.$

b.   $\sqrt{4x^{2}+4x+1}=6$

<=> $\sqrt{(2x+1)^{2}}=6$

<=> $\left | 2x+1 \right |=6$

<=> $\left\{\begin{matrix}2x+1=6 & \\ 2x+1=-6 & \end{matrix}\right.<=>\left\{\begin{matrix}2x=5 & \\ 2x=-7 & \end{matrix}\right.$

<=> $\left\{\begin{matrix}x=\frac{5}{2} & \\ x=\frac{-7}{2} & \end{matrix}\right.$

Vậy $\left\{\begin{matrix}x=\frac{5}{2} & \\ x=\frac{-7}{2} & \end{matrix}\right.$