$2\sqrt{x+3+3\sqrt{2x-3}} - \sqrt{2x-2} = \sqrt{2}(x+1)$
ĐKXĐ : $\begin{cases}2x-3 \geq 0 \\ 2x-2 \geq 0 \\ x+3+3\sqrt{2x-3} \geq0 \end{cases}$
Chia 2 vế cho $\sqrt{2}$
$ PT \Leftrightarrow \sqrt{2x+6+2.3\sqrt{2x-3}} - \sqrt{x-1} = x+1$
$\Leftrightarrow \sqrt{2x-3+2.3\sqrt{2x-3}+9} - \sqrt{x-1} = x+1$
$\Leftrightarrow \sqrt{\left ( \sqrt{2x-3}+3 \right )^2} - \sqrt{x-1}=x+1$
$\Leftrightarrow \sqrt{2x-3}+3-\sqrt{x-1}=x+1$
$\Leftrightarrow \sqrt{2x-3}-\sqrt{x-1}=x-2$
$\Leftrightarrow \frac{x-2}{\sqrt{2x-3}+\sqrt{x-1}}=x-2$
$\Leftrightarrow (x-2)(1-\frac{1}{\sqrt{2x-3}+\sqrt{x-1}})=0$
$TH1 : x-2=0 \Leftrightarrow x=2$
$TH2 : \sqrt{2x-3} + \sqrt{x-1}=1$
$\Leftrightarrow 1-\sqrt{2x-3}=\sqrt{x-1}$
$\Rightarrow 2x-2-2\sqrt{2x-3}=x-1$
$\Leftrightarrow x-1=2\sqrt{2x-3}$
$\Rightarrow x^2-2x+1=4(2x-3)=8x-12$
$\Leftrightarrow x^2-10x+13=0$
$\Leftrightarrow x^2-10x+25-12=0$
$\Leftrightarrow (x-5)^2-12=0 \Leftrightarrow (x-5-\sqrt{12})(x-5+\sqrt{12})=0$
$\Leftrightarrow x=\pm \sqrt{12}+5$
Thử lại : các nghiệm trên thoả PT
Vậy $x=2; \pm \sqrt{12}+5$