3)$x+1+\sqrt{x^{2}-4x+1} =3\sqrt{x} $$\Leftrightarrow (x+1)+\sqrt{(x+1)^2-6x}=3\sqrt{x}$
$\Leftrightarrow 1+\sqrt{1-6(\frac{\sqrt{x}}{x+1})^2}=3(\frac{\sqrt{x}}{x+1})$
Đặt $\frac{\sqrt{x}}{x+1}=a$
$\rightarrow $ PT trở thành: $1+\sqrt{1-6a^2}=3a$
$\rightarrow ...................$