mọi ng làm giúp em bài này nha

[TEX]\sqrt{2x +1} + \sqrt{2x - 3}=\sqrt{x + 3} + sqrt{x-1}[/TEX]
thank mọi người nhìu!!
đk x\geq [TEX] \frac{3}{2}[/TEX]
[TEX]\Leftrightarrow \sqrt{2x +1} - \sqrt{x-1}=\sqrt{x + 3} - \sqrt{2x - 3} [/TEX]
[TEX]\Leftrightarrow ( \sqrt{2x +1} - \sqrt{x-1} )^2=( \sqrt{x + 3} - \sqrt{2x - 3})^2 [/TEX]
[TEX]\Leftrightarrow 3x + 2(\sqrt{2x +1} . \sqrt{x-1})=3x+2( \sqrt{x + 3}. \sqrt{2x - 3}) [/TEX]
[TEX]\Leftrightarrow \sqrt{2x +1}. \sqrt{x-1}=\sqrt{x + 3}. \sqrt{2x - 3} [/TEX]
[TEX]\Leftrightarrow ( \sqrt{2x +1}. \sqrt{x-1})^2=( \sqrt{x + 3}. \sqrt{2x - 3})^2 [/TEX]
\Leftrightarrow[TEX]2x^2-x-1=2x^2+3x-9[/TEX]
[TEX]\Leftrightarrow x= 2[/TEX]