Rút gọn biểu thức A. A.\(A = \frac{{\sqrt x + 1}}{{\sqrt x + 3}}\) B.\(A = \frac{{\sqrt x + 1}}{{\sqrt x - 1}}\) C.\(A = \frac{{\sqrt x - 1}}{{\sqrt x + 1}}\) D.\(A = \frac{{\sqrt x - 1}}{{\sqrt x + 3}}\)
Đáp án đúng: A Giải chi tiết:\(A = \left( {\frac{{2\sqrt x }}{{x - 9}} + \frac{1}{{\sqrt x - 3}}} \right):\frac{3}{{\sqrt x - 3}}\;\;\left( {x \ge 3;\;x \ne 9} \right)\) \(\begin{array}{l}A = \left( {\frac{{2\sqrt x }}{{x - 9}} + \frac{1}{{\sqrt x - 3}}} \right):\frac{3}{{\sqrt x - 3}} = \left( {\frac{{2\sqrt x }}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}} + \frac{1}{{\sqrt x - 3}}} \right).\frac{{\sqrt x - 3}}{3}\\\;\;\; = \frac{{2\sqrt x + \sqrt x + 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\frac{{\left( {\sqrt x - 3} \right)}}{3} = \frac{{3\left( {\sqrt x + 1} \right)}}{{3\left( {\sqrt x - 1} \right)}} = \frac{{\sqrt x + 1}}{{\sqrt x + 3}}.\end{array}\) Chọn A.