Đáp án đúng: A
Giải chi tiết:\(A = \dfrac{{\sqrt {x + 6\sqrt {x - 9} } + \sqrt {x - 6\sqrt {x - 9} } }}{{\sqrt {\dfrac{{81}}{{{x^2}}} - \dfrac{{18}}{x} + 1} }}\), với \(x > 9\)
\(\begin{array}{l}A = \dfrac{{\sqrt {x - 9 + 6\sqrt {x - 9} + 9} + \sqrt {x - 9 - 6\sqrt {x - 9} + 9} }}{{\sqrt {{{\left( {\dfrac{9}{x}} \right)}^2} - 2.\dfrac{9}{x} + 1} }}\,\,\,\,\left( {x > 9} \right)\\A = \dfrac{{\sqrt {{{\left( {\sqrt {x - 9} + 3} \right)}^2}} + \sqrt {{{\left( {\sqrt {x - 9} - 3} \right)}^2}} }}{{\sqrt {{{\left( {\dfrac{9}{x} - 1} \right)}^2}} }}\,\,\,\,\left( {x > 9} \right)\\A = \dfrac{{\left| {\sqrt {x - 9} + 3} \right| + \left| {\sqrt {x - 9} - 3} \right|}}{{\left| {\dfrac{9}{x} - 1} \right|}}\,\,\,\left( {x > 9} \right)\\A = \dfrac{{\sqrt {x - 9} + 3 + \left| {\sqrt {x - 9} - 3} \right|}}{{1 - \dfrac{9}{x}}}\,\,\,\,\,\left( {x > 9 \Rightarrow \dfrac{9}{x} < 1} \right)\end{array}\)
+ TH1: \(\sqrt {x - 9} - 3 \le 0 \Leftrightarrow \sqrt {x - 9} \le 3\, \Leftrightarrow x - 9 \le 9 \Leftrightarrow x \le 18\)
\(\begin{array}{l} \Rightarrow 9 < x \le 18\\ \Rightarrow A = \dfrac{{\sqrt {x - 9} + 3 - \sqrt {x - 9} + 3}}{{\dfrac{{x - 9}}{x}}} = \dfrac{{6x}}{{x - 9}} = 6 + \dfrac{{54}}{{x - 9}} \ge 12.\end{array}\)
Dấu “=” xảy ra \( \Leftrightarrow x = 18\).
+ TH2: \(x > 18\).
\( \Rightarrow A = \dfrac{{\sqrt {x - 9} + 3 + \sqrt {x - 9} - 3}}{{\dfrac{{x - 9}}{x}}} = \dfrac{{2x\sqrt {x - 9} }}{{x - 9}} = \dfrac{{2x}}{{\sqrt {x - 9} }}\).
Ta sẽ chứng minh \(A > 12\), thật vậy:
\(\begin{array}{l}A > 12 \Leftrightarrow \dfrac{{2x}}{{\sqrt {x - 9} }} > 12 \Leftrightarrow 2x > 12.\sqrt {x - 9} \\ \Leftrightarrow 4{x^2} > 144\left( {x - 9} \right) \Leftrightarrow 4{x^2} - 144x + 144.9 > 0\\ \Leftrightarrow 4{\left( {x - 18} \right)^2} > 0\,\,\,\,\left( {do\,x > 18} \right)\end{array}\)
Từ 2 TH ta suy ra \(A \ge 12\,\,\,\,\forall x > 9\).
Vậy GTNN của \(A = 12 \Leftrightarrow x = 18\).
