3. Solve the equation $3 \sqrt{(x+1)^{2}}+\sqrt{x^{2}-4 x+4}=|1+6 x|-4|x-1|$.
Solution: Rewrite the equation as $|3 x+3|+|x-2|+|4 x-4|=|1+6 x|$. Notice that $3 x+3-x+2+4 x-4=1+6 x$, so the solution to the equation is found from the system $\left\{\begin{array}{l}3 x+3 \geq 0, \\ -x+2 \geq 0, \\ 4 x-4 \geq 0, \\ 1+6 x \geq 0 .\end{array} \Rightarrow\left\{\begin{array}{l}x \geq-1, \\ 2 \geq x, \\ x \geq 1, \\ x \geq-1 / 6 .\end{array} \Rightarrow x \in[1,2]\right.\right.$.

Answer: $x \in[1,2]$.