Toán Cho bthức Q=(4√x /”2+√x” -8x/ 4-x):(√x-1/”x-2√x” – 2/√x) Rút Gọn 03/10/2021 By Allison Cho bthức Q=(4√x /”2+√x” -8x/ 4-x):(√x-1/”x-2√x” – 2/√x) Rút Gọn
$$\eqalign{ & Q = \left( {{{4\sqrt x } \over {2 + \sqrt x }} – {{8x} \over {4 – x}}} \right):\left( {{{\sqrt x – 1} \over {x – 2\sqrt x }} – {2 \over {\sqrt x }}} \right)\,\,\left( {x > 0;\,\,x \ne 4} \right) \cr & Q = \left( {{{4\sqrt x } \over {2 + \sqrt x }} – {{8x} \over {\left( {2 + \sqrt x } \right)\left( {2 – \sqrt x } \right)}}} \right):\left( {{{\sqrt x – 1} \over {\sqrt x \left( {\sqrt x – 2} \right)}} – {2 \over {\sqrt x }}} \right) \cr & Q = {{4\sqrt x \left( {2 – \sqrt x } \right) – 8x} \over {\left( {2 + \sqrt x } \right)\left( {2 – \sqrt x } \right)}}:{{\left( {\sqrt x – 1} \right) – 2\left( {\sqrt x – 2} \right)} \over {\sqrt x \left( {\sqrt x – 2} \right)}} \cr & Q = {{8\sqrt x – 4x – 8x} \over {\left( {2 + \sqrt x } \right)\left( {2 – \sqrt x } \right)}}.{{\sqrt x \left( {\sqrt x – 2} \right)} \over {\sqrt x – 1 – 2\sqrt x + 4}} \cr & Q = {{8\sqrt x – 12x} \over {2 + \sqrt x }}{{\sqrt x } \over {3 – \sqrt x }} \cr & Q = {{4\sqrt x \left( {2 – 3\sqrt x } \right)\sqrt x } \over {\left( {2 + \sqrt x } \right)\left( {3 – \sqrt x } \right)}} \cr & Q = {{4x\left( {2 – 3\sqrt x } \right)} \over {\left( {2 + \sqrt x } \right)\left( {3 – \sqrt x } \right)}} \cr} $$ Trả lời
$$\eqalign{
& Q = \left( {{{4\sqrt x } \over {2 + \sqrt x }} – {{8x} \over {4 – x}}} \right):\left( {{{\sqrt x – 1} \over {x – 2\sqrt x }} – {2 \over {\sqrt x }}} \right)\,\,\left( {x > 0;\,\,x \ne 4} \right) \cr
& Q = \left( {{{4\sqrt x } \over {2 + \sqrt x }} – {{8x} \over {\left( {2 + \sqrt x } \right)\left( {2 – \sqrt x } \right)}}} \right):\left( {{{\sqrt x – 1} \over {\sqrt x \left( {\sqrt x – 2} \right)}} – {2 \over {\sqrt x }}} \right) \cr
& Q = {{4\sqrt x \left( {2 – \sqrt x } \right) – 8x} \over {\left( {2 + \sqrt x } \right)\left( {2 – \sqrt x } \right)}}:{{\left( {\sqrt x – 1} \right) – 2\left( {\sqrt x – 2} \right)} \over {\sqrt x \left( {\sqrt x – 2} \right)}} \cr
& Q = {{8\sqrt x – 4x – 8x} \over {\left( {2 + \sqrt x } \right)\left( {2 – \sqrt x } \right)}}.{{\sqrt x \left( {\sqrt x – 2} \right)} \over {\sqrt x – 1 – 2\sqrt x + 4}} \cr
& Q = {{8\sqrt x – 12x} \over {2 + \sqrt x }}{{\sqrt x } \over {3 – \sqrt x }} \cr
& Q = {{4\sqrt x \left( {2 – 3\sqrt x } \right)\sqrt x } \over {\left( {2 + \sqrt x } \right)\left( {3 – \sqrt x } \right)}} \cr
& Q = {{4x\left( {2 – 3\sqrt x } \right)} \over {\left( {2 + \sqrt x } \right)\left( {3 – \sqrt x } \right)}} \cr} $$