Lời giải của Tự Học 365

$P = \left( {\sqrt {ab}  - \dfrac{{ab}}{{a + \sqrt {ab} }}} \right):\dfrac{{\sqrt[4]{{ab}} - \sqrt b }}{{a - b}}$ $= \left( {\dfrac{{\sqrt {ab} \left( {a + \sqrt {ab} } \right) - ab}}{{a + \sqrt {ab} }}} \right).\dfrac{{a - b}}{{\sqrt[4]{{ab}} - {{\left( {\sqrt[4]{b}} \right)}^2}}}$$= \dfrac{{a.\sqrt {ab}  + ab - ab}}{{{{\left( {\sqrt a } \right)}^2} + \sqrt a .\sqrt b }}.\dfrac{{\left( {\sqrt a  - \sqrt b } \right)\left( {\sqrt a  + \sqrt b } \right)}}{{\sqrt[4]{b}\left( {\sqrt[4]{a} - \sqrt[4]{b}} \right)}}$

$= \dfrac{{a\sqrt {ab} }}{{\sqrt a \left( {\sqrt a  + \sqrt b } \right)}}.\dfrac{{\left( {\sqrt a  - \sqrt b } \right)\left( {\sqrt a  + \sqrt b } \right)}}{{\sqrt[4]{b}\left( {\sqrt[4]{a} - \sqrt[4]{b}} \right)}}$  $= \dfrac{{a\sqrt a .\sqrt b }}{{\sqrt a }}.\dfrac{{{{\left( {\sqrt[4]{a}} \right)}^2} - {{\left( {\sqrt[4]{b}} \right)}^2}}}{{\sqrt[4]{b}\left( {\sqrt[4]{a} - \sqrt[4]{b}} \right)}}$ $= \dfrac{{a\sqrt b .\left( {\sqrt[4]{a} - \sqrt[4]{b}} \right)\left( {\sqrt[4]{a} + \sqrt[4]{b}} \right)}}{{\sqrt[4]{b}\left( {\sqrt[4]{a} - \sqrt[4]{b}} \right)}}$ $= \dfrac{{a{{\left( {\sqrt[4]{b}} \right)}^2}.\left( {\sqrt[4]{a} + \sqrt[4]{b}} \right)}}{{\sqrt[4]{b}}} = a\sqrt[4]{b}\left( {\sqrt[4]{a} + \sqrt[4]{b}} \right)$

Vậy $P = a\sqrt[4]{b}(\sqrt[4]{a} + \sqrt[4]{b}).$

Đáp án cần chọn là: d