Cho bthức Q=(4√x /"2+√x" -8x/ 4-x):(√x-1/"x-2√x" - 2/√x) Rút Gọn

$$\eqalign{ & Q = \left( {{{4\sqrt x } \over {2 + \sqrt x }} - {{8x} \over {4 - x}}} \right):\left( {{{\sqrt x - 1} \over {x - 2\sqrt x }} - {2 \over {\sqrt x }}} \right)\,\,\left( {x > 0;\,\,x \ne 4} \right) \cr & Q = \left( {{{4\sqrt x } \over {2 + \sqrt x }} - {{8x} \over {\left( {2 + \sqrt x } \right)\left( {2 - \sqrt x } \right)}}} \right):\left( {{{\sqrt x - 1} \over {\sqrt x \left( {\sqrt x - 2} \right)}} - {2 \over {\sqrt x }}} \right) \cr & Q = {{4\sqrt x \left( {2 - \sqrt x } \right) - 8x} \over {\left( {2 + \sqrt x } \right)\left( {2 - \sqrt x } \right)}}:{{\left( {\sqrt x - 1} \right) - 2\left( {\sqrt x - 2} \right)} \over {\sqrt x \left( {\sqrt x - 2} \right)}} \cr & Q = {{8\sqrt x - 4x - 8x} \over {\left( {2 + \sqrt x } \right)\left( {2 - \sqrt x } \right)}}.{{\sqrt x \left( {\sqrt x - 2} \right)} \over {\sqrt x - 1 - 2\sqrt x + 4}} \cr & Q = {{8\sqrt x - 12x} \over {2 + \sqrt x }}{{\sqrt x } \over {3 - \sqrt x }} \cr & Q = {{4\sqrt x \left( {2 - 3\sqrt x } \right)\sqrt x } \over {\left( {2 + \sqrt x } \right)\left( {3 - \sqrt x } \right)}} \cr & Q = {{4x\left( {2 - 3\sqrt x } \right)} \over {\left( {2 + \sqrt x } \right)\left( {3 - \sqrt x } \right)}} \cr}$$