Cho \(A = \left( {\dfrac{{2x + 1}}{{\sqrt {{x^3}} - 1}} - \dfrac{1}{{\sqrt x - 1}}} \right):\left( {1 - \dfrac{{x + 4}}{{x + \sqrt x + 1}}} \right)\) với \(x \ge 0;\,\,x \ne 1\).
a) Rút gọn A. b) Tìm \(x \in Z\) để \(A \in Z\)
A.\(\begin{array}{l}a)\,\,A = \frac{{\sqrt x }}{{\sqrt x - 3}}\\b)\,\,x \in \left\{ {0;4;16} \right\}\end{array}\)
B.\(\begin{array}{l}a)\,\,A = \frac{{\sqrt x }}{{\sqrt x - 3}}\\b)\,\,x \in \left\{ {0;4;16;36} \right\}\end{array}\)
C.\(\begin{array}{l}a)\,\,A = \frac{{\sqrt x - 3}}{{\sqrt x }}\\b)\,\,x \in \left\{ {0;4;16;36} \right\}\end{array}\)
D.\(\begin{array}{l}a)\,\,A = \frac{{\sqrt x - 3}}{{\sqrt x }}\\b)\,\,x \in \left\{ {4;16;36} \right\}\end{array}\)

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