Tháng Ba 29, 2024

\(R = \left( {\frac{{3\sqrt x }}{{\sqrt x + 2}} + \frac{{\sqrt x }}{{\sqrt x – 2}} + \frac{{3x – 5\sqrt x }}{{4 – x}}} \right):\left( {\frac{{2\sqrt x – 1}}{{\sqrt x – 2}} – 1} \right)\) A \(R= -\frac{{\sqrt x }}{{\sqrt x – 2}}\) B \(R= \frac{{\sqrt x }}{{\sqrt x – 2}}\) C \(R= \frac{{\sqrt x }}{{\sqrt x + 2}}\) D \(R= -\frac{{\sqrt x }}{{\sqrt x + 2}}\)

\(R = \left( {\frac{{3\sqrt x }}{{\sqrt x + 2}} + \frac{{\sqrt x }}{{\sqrt x – 2}} + \frac{{3x – 5\sqrt x }}{{4 – x}}} \right):\left( {\frac{{2\sqrt x – 1}}{{\sqrt x – 2}} – 1} \right)\)

A \(R= -\frac{{\sqrt x }}{{\sqrt x – 2}}\)

B \(R= \frac{{\sqrt x }}{{\sqrt x – 2}}\)

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

D \(R= -\frac{{\sqrt x }}{{\sqrt x + 2}}\)

Hướng dẫn Chọn đáp án là: C

Lời giải chi tiết:

\(R = \left( {\frac{{3\sqrt x }}{{\sqrt x + 2}} + \frac{{\sqrt x }}{{\sqrt x – 2}} + \frac{{3x – 5\sqrt x }}{{4 – x}}} \right):\left( {\frac{{2\sqrt x – 1}}{{\sqrt x – 2}} – 1} \right)\)

Điều kiện: \(x \ge 0,\,\,\,x \ne 4.\)

\(\begin{array}{l}R = \left( {\frac{{3\sqrt x }}{{\sqrt x + 2}} + \frac{{\sqrt x }}{{\sqrt x – 2}} + \frac{{3x – 5\sqrt x }}{{4 – x}}} \right):\left( {\frac{{2\sqrt x – 1}}{{\sqrt x – 2}} – 1} \right)\\\,\,\,\, = \left( {\frac{{3\sqrt x }}{{\sqrt x + 2}} + \frac{{\sqrt x }}{{\sqrt x – 2}} – \frac{{3x – 5\sqrt x }}{{(\sqrt x – 2)(\sqrt x + 2)}}} \right):\left( {\frac{{2\sqrt x – 1 – \sqrt x + 2}}{{\sqrt x – 2}}} \right)\\\,\,\,\, = \frac{{3\sqrt x \left( {\sqrt x – 2} \right) + \sqrt x \left( {\sqrt x + 2} \right) – 3x + 5\sqrt x }}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}:\frac{{\sqrt x + 1}}{{\sqrt x – 2}}\\\,\,\,\, = \frac{{3x – 6\sqrt x + x + 2\sqrt x – 3x + 5\sqrt x }}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}:\frac{{\sqrt x + 1}}{{\sqrt x – 2}}\\\,\,\, = \frac{{x + \sqrt x }}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}:\frac{{\sqrt x + 1}}{{\sqrt x – 2}}\\\,\,\, = \frac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}.\frac{{\sqrt x – 2}}{{\sqrt x + 1}}\\\,\,\, = \frac{{\sqrt x }}{{\sqrt x + 2}}.\end{array}\)