Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = \left( {2\sqrt 3 - 5\sqrt {27} + 4\sqrt {12} } \right):\sqrt 3 - \sqrt[3]{{64}}\\
= \left( {2\sqrt 3 - 5\sqrt {{3^2}.3} + 4\sqrt {{2^2}.3} } \right):\sqrt 3 - \sqrt[3]{{{4^3}}}\\
= \left( {2\sqrt 3 - 5.3\sqrt 3 + 4.2.\sqrt 3 } \right):\sqrt 3 - 4\\
= \left( {2\sqrt 3 - 15\sqrt 3 + 8\sqrt 3 } \right):\sqrt 3 - 4\\
= \left( { - 5\sqrt 3 } \right):\sqrt 3 - 4\\
= - 5 - 4\\
= - 9\\
B = \left( {1 - \dfrac{{x + \sqrt x }}{{1 + \sqrt x }}} \right)\left( {1 + \dfrac{{x + \sqrt x }}{{1 - \sqrt x }}} \right)\\
= \left( {1 - \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{1 + \sqrt x }}} \right)\dfrac{{\left( {1 - \sqrt x } \right) + x + \sqrt x }}{{1 - \sqrt x }}\\
= \left( {1 - \sqrt x } \right).\dfrac{{x + 1}}{{1 - \sqrt x }}\\
= x + 1\\
b,\\
B > A \Leftrightarrow x + 1 > - 9 \Leftrightarrow x > - 10\,\,\,\,\,\,\left( {\forall x \ge 0} \right)\\
\Rightarrow x \ge 0,x \ne 1
\end{array}\)
