by Rút gọn biểu thức: \(B=\frac{1}{\sqrt{x}+1}-{{\frac{x+2}{x\sqrt{x}+1}}^{{}}}{{^{{}}}^{{}}}\left( x\ge 0 \right)\)
A \(\frac{{ – 2}}{{x – \sqrt x + 1}}\)
B \(\frac{{ – 1}}{{x – \sqrt x + 1}}\)
C \(\frac{{ 1}}{{x – \sqrt x + 1}}\)
D \(\frac{{ – 3}}{{x – \sqrt x + 1}}\)
Hướng dẫn Chọn đáp án là: B
Lời giải chi tiết:
\(\begin{array}{l}B = \frac{1}{{\sqrt x + 1}} – \frac{{x + 2}}{{x\sqrt x + 1}} = \frac{1}{{\sqrt x + 1}} – \frac{{x + 2}}{{\left( {\sqrt x + 1} \right)\left( {x – \sqrt x + 1} \right)}}\\ = \frac{{x – \sqrt x + 1 – x – 2}}{{\left( {\sqrt x + 1} \right)\left( {x – \sqrt x + 1} \right)}} = \frac{{ – \sqrt x – 1}}{{\left( {\sqrt x + 1} \right)\left( {x – \sqrt x + 1} \right)}}\\ = \frac{{ – \left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {x – \sqrt x + 1} \right)}} = \frac{{ – 1}}{{x – \sqrt x + 1}}\end{array}\)
