Đáp án:
b) \(\dfrac{{ - 4 + 3\sqrt 3 }}{{11}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne 1\\
A = \dfrac{{2x + 1 - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}:\dfrac{{x + \sqrt x + 1 - x + 2}}{{x + \sqrt x + 1}}\\
= \dfrac{{x - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{x + \sqrt x + 1}}{{\sqrt x + 3}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 3}}\\
b)Thay:x = \dfrac{{2 - \sqrt 3 }}{2} = \dfrac{{4 - 2\sqrt 3 }}{4} = \dfrac{{3 - 2\sqrt 3 .1 + 1}}{4}\\
= \dfrac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{4}\\
\to A = \dfrac{{\sqrt {\dfrac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{4}} }}{{\sqrt {\dfrac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{4}} + 3}} = \dfrac{{\dfrac{{\sqrt 3 - 1}}{2}}}{{\dfrac{{\sqrt 3 - 1}}{2} + 3}} = \dfrac{{\sqrt 3 - 1}}{2}:\dfrac{{\sqrt 3 - 1 + 6}}{2}\\
= \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 5}} = \dfrac{{ - 4 + 3\sqrt 3 }}{{11}}
\end{array}\)
