Giải hộ mình câu này với các bạn

ĐKXĐ
$\displaystyle \begin{cases}
x^{2} -3x+2 & \geq 0\\
x+3 & \geq 0\\
x^{2} +2x-3 & \geq 0\\
x-2 & \geq 0
\end{cases} \Rightarrow \begin{cases}
( x-2)( x-1) & \geq 0\\
x & \geq -3\\
( x-1)( x+3) & \geq 0\\
x & \geq 2
\end{cases} \Rightarrow \begin{cases}
\left[ \begin{array}{l l}
x & \geq 2\\
x & \leq 1
\end{array} \right.\\
x\geq -3\\
\left[ \begin{array}{l l}
x & \geq 1\\
x & \leq \ -3
\end{array} \right.\\
x\geq 2
\end{cases} \Rightarrow \left[ \begin{array}{l l}
x\geq 2\\
-3\leq x\leq 1
\end{array} \right.$
Ta có
$\displaystyle \sqrt{x^{2} -3x+2} +\sqrt{x+3} =\sqrt{x^{2} +2x-3} +\sqrt{x-2}$
$\displaystyle \sqrt{( x-2)( x-1)} +\sqrt{x+3} =\sqrt{( x-1)( x+3)} +\sqrt{x-2}$
$\displaystyle \sqrt{( x-2)( x-1)} -\sqrt{( x-1)( x+3)} =\sqrt{x-2} -\sqrt{x+3}$
$\displaystyle \sqrt{x-1} \times \left(\sqrt{x-2} -\sqrt{x+3}\right) -\left(\sqrt{x-2} -\sqrt{x+3}\right) =0$
$\displaystyle \left(\sqrt{x-2} -\sqrt{x+3}\right) \times \left(\sqrt{x-1} -1\right) =0$
$\displaystyle \left[ \begin{array}{l l}
\sqrt{x-2} -\sqrt{x+3} & =0\\
\sqrt{x-1} -1 & =0
\end{array} \right. \Rightarrow \left[ \begin{array}{l l}
\sqrt{x-2} & =\sqrt{x+3}\\
\sqrt{x-1} & =1
\end{array} \right. \Rightarrow \left[ \begin{array}{l l}
\left(\sqrt{x-2}\right)^{2} & =\left(\sqrt{x+3}\right)^{2}\\
\left(\sqrt{x-1}\right)^{2} & =1^{2}
\end{array} \right.$
$\displaystyle \Rightarrow \left[ \begin{array}{l l}
x-2 & =x+3\\
x-1 & =1
\end{array} \right. \Rightarrow \left[ \begin{array}{l l}
0x & =5\ ( vô\ lý)\\
x & =2
\end{array} \right. \Rightarrow x=2$ thỏa mãn ĐKXĐ
Vậy$\displaystyle \ x=2$