Đáp án:
a. \(\dfrac{{x + 1}}{{4\sqrt x }}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x > 0;x \ne 1\\
P = \left[ {\dfrac{{x + 2\sqrt x + 1 + x - \sqrt x - x - \sqrt x }}{{x - 1}}} \right]:\left[ {\dfrac{{x + 2\sqrt x + 1 - x + 2\sqrt x - 1}}{{x - 1}}} \right]\\
= \dfrac{{x + 1}}{{x - 1}}.\dfrac{{x - 1}}{{4\sqrt x }}\\
= \dfrac{{x + 1}}{{4\sqrt x }}\\
b.Thay:x = \dfrac{{2 - \sqrt 3 }}{2} = \dfrac{{4 - 2\sqrt 3 }}{4}\\
= \dfrac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{4}\\
P = \dfrac{{\dfrac{{2 - \sqrt 3 }}{2} + 1}}{{4\sqrt {\dfrac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{4}} }} = \left( {\dfrac{{2 - \sqrt 3 }}{2} + 1} \right):\left( {4.\dfrac{{\sqrt 3 - 1}}{2}} \right)\\
= \left( {\dfrac{{2 - \sqrt 3 + 2}}{2}} \right):\left( {2\sqrt 3 - 2} \right)\\
= \dfrac{{2 - \sqrt 3 + 2}}{{2\left( {2\sqrt 3 - 2} \right)}}\\
= \dfrac{{4 - \sqrt 3 }}{{4\sqrt 3 - 4}}
\end{array}\)
