by \(Q = \left( {\frac{{x + 2}}{{\sqrt x + 1}} – \sqrt x } \right):\left( {\frac{{\sqrt x – 4}}{{1 – x}} – \frac{{\sqrt x }}{{\sqrt x + 1}}} \right)\)
A \(Q= \frac{{\sqrt x + 1}}{{\sqrt x – 2}}\)
B \(Q= \frac{{\sqrt x – 1}}{{\sqrt x – 2}}\)
C \(Q= \frac{{\sqrt x + 1}}{{\sqrt x + 2}}\)
D \(Q= \frac{{\sqrt x – 1}}{{\sqrt x + 2}}\)
Hướng dẫn Chọn đáp án là: D
Lời giải chi tiết:
\(Q = \left( {\frac{{x + 2}}{{\sqrt x + 1}} – \sqrt x } \right):\left( {\frac{{\sqrt x – 4}}{{1 – x}} – \frac{{\sqrt x }}{{\sqrt x + 1}}} \right)\)
Điều kiện: \(x \ge 0,\,\,x \ne 1,\,\,x \ne 4.\)
\(\begin{array}{l}Q = \left( {\frac{{x + 2}}{{\sqrt x + 1}} – \sqrt x } \right):\left( {\frac{{\sqrt x – 4}}{{1 – x}} – \frac{{\sqrt x }}{{\sqrt x + 1}}} \right)\\\,\,\,\,\, = \left( {\frac{{x + 2 – \sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}} \right):\left( {\frac{{4 – \sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x – 1} \right)}} – \frac{{\sqrt x }}{{\sqrt x + 1}}} \right)\\\,\,\,\, = \frac{{x + 2 – x – \sqrt x }}{{\sqrt x + 1}}:\frac{{4 – \sqrt x – \sqrt x \left( {\sqrt x – 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x – 1} \right)}}\\\,\,\,\, = \frac{{2 – \sqrt x }}{{\sqrt x + 1}}:\frac{{4 – \sqrt x – x + \sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x – 1} \right)}} = \frac{{2 – \sqrt x }}{{\sqrt x + 1}}:\frac{{4 – x}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x – 1} \right)}}\\\,\,\, = \frac{{2 – \sqrt x }}{{\sqrt x + 1}}.\frac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x – 1} \right)}}{{\left( {2 – \sqrt x } \right)\left( {2 + \sqrt x } \right)}}\, = \frac{{\sqrt x – 1}}{{\sqrt x + 2}}.\end{array}\)
