Đáp án:

\(\begin{array}{l}
a)\dfrac{{x + \sqrt x  + 1}}{{\sqrt x }}\\
b)\dfrac{{\sqrt x  - 1}}{{\sqrt x  + 1}}\\
c)\dfrac{{{{\left( {\sqrt x  + 1} \right)}^2}}}{{\sqrt x }}\\
d)\dfrac{{x - 1}}{{\sqrt x }}
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)P = \dfrac{{\sqrt x  + 1 + x}}{{\sqrt x \left( {\sqrt x  + 1} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x  + 1} \right)}}{{\sqrt x }}\\
 = \dfrac{{x + \sqrt x  + 1}}{{\sqrt x }}\\
b)P = \dfrac{{\sqrt x \left( {\sqrt x  + 1} \right) + 3\left( {\sqrt x  - 1} \right) - 6\sqrt x  + 4}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{x + \sqrt x  + 3\sqrt x  - 3 - 6\sqrt x  + 4}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{x - 2\sqrt x  + 1}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{{{\left( {\sqrt x  - 1} \right)}^2}}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 1}}\\
c)P = \dfrac{{x - 1}}{{\sqrt x }}:\dfrac{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right) + 1 - \sqrt x }}{{\sqrt x \left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{x - 1}}{{\sqrt x }}.\dfrac{{\sqrt x \left( {\sqrt x  + 1} \right)}}{{x - 1 + 1 - \sqrt x }}\\
 = \dfrac{{x - 1}}{{\sqrt x }}.\dfrac{{\sqrt x \left( {\sqrt x  + 1} \right)}}{{\sqrt x \left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{{{\left( {\sqrt x  + 1} \right)}^2}}}{{\sqrt x }}\\
d)P = \dfrac{{x - 1}}{{\sqrt x \left( {\sqrt x  - 1} \right)}}:\dfrac{{\sqrt x  - 1 + 2}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}{{\sqrt x \left( {\sqrt x  - 1} \right)}}.\dfrac{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}{{\sqrt x  + 1}}\\
 = \dfrac{{x - 1}}{{\sqrt x }}
\end{array}\)