Rút gọn biểu thức sau: \(A = \left( {1 - \dfrac{{2\sqrt x }}{{3\sqrt x + 1}} - \dfrac{{1 - 2\sqrt x }}{{1 - 9x}}} \right):\left( {\dfrac{{6\sqrt x + 5}}{{3\sqrt x + 1}} - 2} \ri...

* Hướng dẫn giải

\(\begin{array}{l}A = \left( {1 - \dfrac{{2\sqrt x }}{{3\sqrt x  + 1}} - \dfrac{{1 - 2\sqrt x }}{{1 - 9x}}} \right):\left( {\dfrac{{6\sqrt x  + 5}}{{3\sqrt x  + 1}} - 2} \right)\;\;\;\left( {x \ge 0,\;\;x \ne \dfrac{1}{9}} \right)\\\;\;\; = \left( {1 - \dfrac{{2\sqrt x }}{{3\sqrt x  + 1}} + \dfrac{{1 - 2\sqrt x }}{{\left( {3\sqrt x  + 1} \right)\left( {3\sqrt x  - 1} \right)}}} \right):\left( {\dfrac{{6\sqrt x  + 5 - 2\left( {3\sqrt x  + 1} \right)}}{{3\sqrt x  + 1}}} \right)\\\;\;\; = \dfrac{{9x - 1 - 2\sqrt x \left( {3\sqrt x  - 1} \right) + 1 - 2\sqrt x }}{{\left( {3\sqrt x  + 1} \right)\left( {3\sqrt x  - 1} \right)}}:\dfrac{{6\sqrt x  + 5 - 6\sqrt x  - 2}}{{3\sqrt x  + 1}}\\\;\;\; = \dfrac{{9x - 1 - 6x + 2\sqrt x  + 1 - 2\sqrt x }}{{\left( {3\sqrt x  + 1} \right)\left( {3\sqrt x  - 1} \right)}}.\dfrac{{3\sqrt x  + 1}}{3}\\\;\;\; = \dfrac{{3x}}{{3\left( {3\sqrt x  - 1} \right)}} = \dfrac{x}{{3\sqrt x  - 1}}.\end{array}\)