Đáp án:
a. \(\dfrac{{\sqrt x - 4}}{{\sqrt x + 4}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne 16\\
P = \dfrac{{\sqrt x + 3}}{{\sqrt x + 4}} + \dfrac{{\sqrt x - 2}}{{\sqrt x - 4}} - \dfrac{{x + 9\sqrt x - 36}}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)}}\\
= \dfrac{{x - \sqrt x - 12 + x + 2\sqrt x - 8 - x - 9\sqrt x + 36}}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)}}\\
= \dfrac{{x - 8\sqrt x + 16}}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)}}\\
= \dfrac{{{{\left( {\sqrt x - 4} \right)}^2}}}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)}}\\
= \dfrac{{\sqrt x - 4}}{{\sqrt x + 4}}\\
b.x = 3 - 2\sqrt x \\
\to x + 2\sqrt x - 3 = 0\\
\to \left[ \begin{array}{l}
\sqrt x = 1\\
\sqrt x = - 3\left( l \right)
\end{array} \right.\\
\to x = 1\\
Thay:x = 1\\
\to P = \dfrac{{1 - 4}}{{1 + 4}} = - \dfrac{3}{5}\\
c.P > \dfrac{1}{3}\\
\to \dfrac{{\sqrt x - 4}}{{\sqrt x + 4}} > \dfrac{1}{3}\\
\to \dfrac{{3\sqrt x - 12 - \sqrt x - 4}}{{3\left( {\sqrt x + 4} \right)}} > 0\\
\to 2\sqrt x - 16 > 0\left( {do:\sqrt x + 4 > 0\forall x \ge 0} \right)\\
\to \sqrt x - 8 > 0\\
\to x > 64
\end{array}\)
