Cho \(A = \left( {1 - \dfrac{{\sqrt x }}{{1 + \sqrt x }}} \right):\left( {\dfrac{{\sqrt x + 3}}{{\sqrt x - 2}} + \dfrac{{\sqrt x + 2}}{{3 - \sqrt x }} + \dfrac{{\sqrt x + 2}}{{x - 5\sqrt x + 6}}} \right)\) với \(x \ge 0,x \ne 4,x \ne 9.\)
a) Rút gọn A.
b) Tìm \(x \in Z\) để \(A \in Z\)
c) Tìm x để \(A < 0.\)
A.\(\begin{array}{l}a)\,\,A = \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}}\\b)\,\,x \in \left\{ {0} \right\}\\c)\,\,0 \le x < 4\end{array}\)
B.\(\begin{array}{l}a)\,\,A = \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}}\\b)\,\,x \in \left\{ {0;4} \right\}\\c)\,\,0 < x < 4\end{array}\)
C.\(\begin{array}{l}a)\,\,A = \dfrac{3}{{\sqrt x + 1}}\\b)\,\,x \in \left\{ {0;4} \right\}\\c)\,\,0 \le x < 4\end{array}\)
D.\(\begin{array}{l}a)\,\,A = \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}}\\b)\,\,x \in \left\{ {0; \pm 4} \right\}\\c)\,\,0 \le x < 4\end{array}\)
a) Rút gọn A.
b) Tìm \(x \in Z\) để \(A \in Z\)
c) Tìm x để \(A < 0.\)
A.\(\begin{array}{l}a)\,\,A = \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}}\\b)\,\,x \in \left\{ {0} \right\}\\c)\,\,0 \le x < 4\end{array}\)
B.\(\begin{array}{l}a)\,\,A = \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}}\\b)\,\,x \in \left\{ {0;4} \right\}\\c)\,\,0 < x < 4\end{array}\)
C.\(\begin{array}{l}a)\,\,A = \dfrac{3}{{\sqrt x + 1}}\\b)\,\,x \in \left\{ {0;4} \right\}\\c)\,\,0 \le x < 4\end{array}\)
D.\(\begin{array}{l}a)\,\,A = \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}}\\b)\,\,x \in \left\{ {0; \pm 4} \right\}\\c)\,\,0 \le x < 4\end{array}\)
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Hóa Học lớp 12
