Đáp án:
\(\dfrac{{\sqrt x + 6}}{{\sqrt x - 2}}\)
Giải thích các bước giải:
\(\begin{array}{l}
P = \dfrac{{\sqrt x \left( {\sqrt x + 2} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}} + \dfrac{{x + 12}}{{x - 4}} - \dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{\sqrt x \left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x - 2}} + \dfrac{{x + 12}}{{x - 4}} - \dfrac{{\sqrt x - 2}}{{\sqrt x + 2}}\\
= \dfrac{{{{\left( {\sqrt x + 2} \right)}^2} + x + 12 - {{\left( {\sqrt x - 2} \right)}^2}}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x + 4\sqrt x + 4 + x + 12 - x + 4\sqrt x - 4}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x + 8\sqrt x + 12}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\left( {\sqrt x + 6} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x + 6}}{{\sqrt x - 2}}
\end{array}\)
