Cho \(A = \left( {\dfrac{1}{{\sqrt x + 1}} - \dfrac{{2\sqrt x - 2}}{{x\sqrt x - \sqrt x + x - 1}}} \right):\left( {\dfrac{1}{{\sqrt x - 1}} - \dfrac{2}{{x - 1}}} \right)\) với \(x \ge 0,x \ne 1.\)
a) Rút gọn A.
b) Tìm\(x \in Z\) để \(A \in Z\)
c) Tìm x để A đạt GTNN.
A.\(\begin{array}{l}a)\,\,A = \dfrac{{ - 2}}{{\sqrt x + 1}}\\b)\,\,x \in \left\{ {0;2} \right\}\\c)\,\,\min A = - 1\end{array}\)
B.\(\begin{array}{l}a)\,\,A = \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\b)\,\,x = 0\\c)\,\,\min A = - 1\end{array}\)
C.\(\begin{array}{l}a)\,\,A = \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\b)\,\,x \in \left\{ {0;\pm2} \right\}\\c)\,\,\min A = 1\end{array}\)
D.\(\begin{array}{l}a)\,\,A = \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\b)\,\,x \in \left\{ {0;2} \right\}\\c)\,\,\min A = - 1\end{array}\)
a) Rút gọn A.
b) Tìm\(x \in Z\) để \(A \in Z\)
c) Tìm x để A đạt GTNN.
A.\(\begin{array}{l}a)\,\,A = \dfrac{{ - 2}}{{\sqrt x + 1}}\\b)\,\,x \in \left\{ {0;2} \right\}\\c)\,\,\min A = - 1\end{array}\)
B.\(\begin{array}{l}a)\,\,A = \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\b)\,\,x = 0\\c)\,\,\min A = - 1\end{array}\)
C.\(\begin{array}{l}a)\,\,A = \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\b)\,\,x \in \left\{ {0;\pm2} \right\}\\c)\,\,\min A = 1\end{array}\)
D.\(\begin{array}{l}a)\,\,A = \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\b)\,\,x \in \left\{ {0;2} \right\}\\c)\,\,\min A = - 1\end{array}\)
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