Tháng Tư 19, 2024

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

\(P = \left( {\frac{2}{{\sqrt x – 2}} + \frac{{\sqrt x – 1}}{{2\sqrt x – x}}} \right):\left( {\frac{{\sqrt x + 2}}{{\sqrt x }} – \frac{{\sqrt x – 1}}{{\sqrt x – 2}}} \right)\)

A \(P= \frac{{\sqrt x + 1}}{{\sqrt x – 4}}\)

B \(P= \frac{{\sqrt x – 1}}{{\sqrt x – 4}}\)

C \(P= \frac{{\sqrt x + 1}}{{\sqrt x + 4}}\)

D \(P= \frac{{\sqrt x – 1}}{{\sqrt x + 4}}\)

Hướng dẫn Chọn đáp án là: A

Lời giải chi tiết:

\(P = \left( {\frac{2}{{\sqrt x – 2}} + \frac{{\sqrt x – 1}}{{2\sqrt x – x}}} \right):\left( {\frac{{\sqrt x + 2}}{{\sqrt x }} – \frac{{\sqrt x – 1}}{{\sqrt x – 2}}} \right)\)

Điều kiện: \(x > 0;x \ne 4\)

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