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Giải thích các bước giải:

ĐKXĐ: `x>0; x\ne4; x\ne9`

$\begin{array}{l} \left( {\dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} - \dfrac{{2\sqrt x }}{{\sqrt x + 2}} + \dfrac{{5\sqrt x + 2}}{{4 - x}}} \right):\dfrac{{3\sqrt x - x}}{{x + 4\sqrt x + 4}}\\ = \left( {\dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} - \dfrac{{2\sqrt x }}{{\sqrt x + 2}} - \dfrac{{5\sqrt x + 2}}{{x - 4}}} \right):\dfrac{{3\sqrt x - x}}{{{{\left( {\sqrt x + 2} \right)}^2}}}\\ = = \left( {\dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} - \dfrac{{2\sqrt x }}{{\sqrt x + 2}} - \dfrac{{5\sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}} \right):\dfrac{{3\sqrt x - x}}{{{{\left( {\sqrt x + 2} \right)}^2}}}\\ = \left( {\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 2} \right) - 2\sqrt x \left( {\sqrt x - 2} \right) - \left( {5\sqrt x + 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}} \right).\dfrac{{{{\left( {\sqrt x + 2} \right)}^2}}}{{3\sqrt x - x}}\\ = \left( {\dfrac{{x + 2\sqrt x + \sqrt x + 2 - 2x + 4\sqrt x - 5\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}} \right).\dfrac{{{{\left( {\sqrt x + 2} \right)}^2}}}{{\sqrt x \left( {3 - \sqrt x } \right)}}\\ = \left( {\dfrac{{ - x + 2\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}} \right).\dfrac{{{{\left( {\sqrt x + 2} \right)}^2}}}{{\sqrt x \left( {3 - \sqrt x } \right)}}\\ = \left( {\dfrac{{ - \sqrt x \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}} \right).\dfrac{{{{\left( {\sqrt x + 2} \right)}^2}}}{{\sqrt x \left( {3 - \sqrt x } \right)}}\\ = \dfrac{{ - \left( {\sqrt x + 2} \right)}}{{3 - \sqrt x }}\\ = \dfrac{{\sqrt x + 2}}{{\sqrt x - 3}} \end{array}$