\(S = \left( {\frac{{\sqrt x + 3}}{{\sqrt x – 2}} + \frac{{\sqrt x + 2}}{{3 – \sqrt x }} + \frac{{\sqrt x + 2}}{{x – 5\sqrt x + 6}}} \right):\left( {1 – \frac{{\sqrt x }}{{\sqrt x + 1}}} \right)\) A \(S= \frac{{\sqrt x – 1}}{{\sqrt x – 2}}\) B \(S= \frac{{\sqrt x + 1}}{{\sqrt x – 2}}\) C \(S= \frac{{\sqrt x + 1}}{{\sqrt x + 2}}\) D \(S= \frac{{\sqrt x – 1}}{{\sqrt x + 2}}\)

\(S = \left( {\frac{{\sqrt x + 3}}{{\sqrt x – 2}} + \frac{{\sqrt x + 2}}{{3 – \sqrt x }} + \frac{{\sqrt x + 2}}{{x – 5\sqrt x + 6}}} \right):\left( {1 – \frac{{\sqrt x }}{{\sqrt x + 1}}} \right)\)

A \(S= \frac{{\sqrt x – 1}}{{\sqrt x – 2}}\)

B \(S= \frac{{\sqrt x + 1}}{{\sqrt x – 2}}\)

C \(S= \frac{{\sqrt x + 1}}{{\sqrt x + 2}}\)

D \(S= \frac{{\sqrt x – 1}}{{\sqrt x + 2}}\)

Hướng dẫn Chọn đáp án là: B

Lời giải chi tiết:

\(\begin{array}{l}4)\,\,S = \left( {\frac{{\sqrt x + 3}}{{\sqrt x – 2}} + \frac{{\sqrt x + 2}}{{3 – \sqrt x }} + \frac{{\sqrt x + 2}}{{x – 5\sqrt x + 6}}} \right):\left( {1 – \frac{{\sqrt x }}{{\sqrt x + 1}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {DK:\,\,x \ge 0,\,\,x \ne 4,\,\,x \ne 9} \right)\\\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{\sqrt x + 3}}{{\sqrt x – 2}} – \frac{{\sqrt x + 2}}{{\sqrt x – 3}} + \frac{{\sqrt x + 2}}{{(\sqrt x – 3).(\sqrt x – 2)}}} \right):\left( {\frac{{\sqrt x + 1 – \sqrt x }}{{\sqrt x + 1}}} \right)\\\,\,\,\,\,\,\,\,\,\, = \frac{{(\sqrt x + 3)(\sqrt x – 3) – (\sqrt x + 2)(\sqrt x – 2) + \sqrt x + 2}}{{(\sqrt x – 3).(\sqrt x – 2)}}:\frac{1}{{\sqrt x + 1}}\\\,\,\,\,\,\,\,\,\,\, = \frac{{x – 9 – (x – 4) + \sqrt x + 2}}{{(\sqrt x – 3).(\sqrt x – 2)}}:\frac{1}{{\sqrt x + 1}}\\\,\,\,\,\,\,\,\,\, = \frac{{x – 9 – x + 4 + \sqrt x + 2}}{{(\sqrt x – 3).(\sqrt x – 2)}}:\frac{1}{{\sqrt x + 1}}\\\,\,\,\,\,\,\,\, = \frac{{\sqrt x – 3}}{{(\sqrt x – 3).(\sqrt x – 2)}}.(\sqrt x + 1)\\\,\,\,\,\,\,\,\, = \frac{{\sqrt x + 1}}{{\sqrt x – 2}}.\end{array}\)