Giải thích các bước giải:
$x^2+2\sqrt{x^2-3x+1}=3x+4$
$\to (x^2-3x+1)+2\sqrt{x^2-3x+1}+1=6$
$\to (\sqrt{x^2-3x+1}+1)^2=6$
$\to \sqrt{x^2-3x+1}+1=\sqrt{6}$
$\to \sqrt{x^2-3x+1}=\sqrt{6}-1$
$\to x^2-3x+1=(\sqrt{6}-1)^2=7-2\sqrt{6}$
$\to x^2-3x+2\sqrt{6}-6=0$
$\to x=\dfrac{3+\sqrt{33-8\sqrt{6}}}{2},\:x=\dfrac{3-\sqrt{33-8\sqrt{6}}}{2}$