$\sqrt{\frac{6}{3-x}}+\sqrt{\frac{8}{2-x}}=6$ giải giúp mình nha mọi người
[TEX]\huge \blue x < 2 [/TEX]
[TEX]\huge \blue \sqrt{\frac{6}{3-x}}-2+\sqrt{\frac{8}{2-x}}-4=0 \Leftrightarrow \frac{\frac{4x-6}{3-x}}{\sqrt{\frac{6}{3-x}}+2}+\frac{\frac{8(2x-3)}{2-x}}{\sqrt{\frac{8}{2-x}}+4}=0\Leftrightarrow x=\frac{2}{3}[/TEX]