Cho biểu thức: \(A = \frac{{3x + \sqrt {16x} - 7}}{{x - \sqrt {4x} - 3}} + \frac{{1 - \sqrt x }}{{\sqrt x - 3}} - \frac{{\sqrt x + 3}}{{\sqrt x + 1}}(x \ge 0,x \ne 9).\)
a) Rút gọn A.
b) Tìm x nguyên để A nguyên.
A.\(\begin{array}{l}a)\,A = \frac{{\sqrt x - 3}}{{\sqrt x + 3}}\\b)\,x \in \left\{ {0;\,1;\,4;\,16;\,25;\,36;\,81} \right\}\end{array}\)
B.\(\begin{array}{l}a)\,A = \frac{{\sqrt x + 3}}{{\sqrt x - 3}}\\b)\,x \in \left\{ {16;\,25;\,36;\,81} \right\}\end{array}\)
C.\(\begin{array}{l}a)\,A = \frac{{\sqrt x + 3}}{{\sqrt x - 3}}\\b)\,x \in \left\{ {0;\,1;\,4;\,16;\,25;\,36;\,81} \right\}\end{array}\)
D.\(\begin{array}{l}a)\,A = \frac{{2\sqrt x + 3}}{{\sqrt x - 3}}\\b)\,x \in \left\{ {0;\,1;\,4;\,16;\,25;\,36;\,81} \right\}\end{array}\)
a) Rút gọn A.
b) Tìm x nguyên để A nguyên.
A.\(\begin{array}{l}a)\,A = \frac{{\sqrt x - 3}}{{\sqrt x + 3}}\\b)\,x \in \left\{ {0;\,1;\,4;\,16;\,25;\,36;\,81} \right\}\end{array}\)
B.\(\begin{array}{l}a)\,A = \frac{{\sqrt x + 3}}{{\sqrt x - 3}}\\b)\,x \in \left\{ {16;\,25;\,36;\,81} \right\}\end{array}\)
C.\(\begin{array}{l}a)\,A = \frac{{\sqrt x + 3}}{{\sqrt x - 3}}\\b)\,x \in \left\{ {0;\,1;\,4;\,16;\,25;\,36;\,81} \right\}\end{array}\)
D.\(\begin{array}{l}a)\,A = \frac{{2\sqrt x + 3}}{{\sqrt x - 3}}\\b)\,x \in \left\{ {0;\,1;\,4;\,16;\,25;\,36;\,81} \right\}\end{array}\)
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Hóa Học lớp 12
