a) Điều kiện: \(x \ge 0,x \ne 4\)
Ta có:
\(P\) = \(\dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} + \dfrac{{2\sqrt x }}{{\sqrt x + 2}} + \dfrac{{2 + 5\sqrt x }}{{4 - x}}\)
\( = \dfrac{{(\sqrt x + 1)(\sqrt x + 2)}}{{{{(\sqrt x )}^2} - {2^2}}} \)\(+ \dfrac{{2\sqrt x (\sqrt x - 2)}}{{{{(\sqrt x )}^2} - {2^2}}} - \dfrac{{2 + 5\sqrt x }}{{x - 4}}\)
\( = \dfrac{{x + 2\sqrt x + \sqrt x + 2}}{{x - 4}} + \dfrac{{2x - 4\sqrt x }}{{x - 4}}\)\( - \dfrac{{2 + 5\sqrt x }}{{x - 4}}\)
\( = \dfrac{{x + 3\sqrt x + 2 + 2x - 4\sqrt x - 2 - 5\sqrt x }}{{x - 4}}\)
\( = \dfrac{{3x - 6\sqrt x }}{{x - 4}} \)\(= \dfrac{{3\sqrt x (\sqrt x - 2)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}} \)\(= \dfrac{{3\sqrt x }}{{\sqrt x + 2}}\)
b) Ta có: \(P = 2\) (ĐK:\(x \ge 0,x \ne 4\))
\(\eqalign{
& \Leftrightarrow {{3\sqrt x } \over {\sqrt x + 2}} = 2 \cr
& \Leftrightarrow 3\sqrt x = 2(\sqrt x + 2) \cr
&\Leftrightarrow 3\sqrt x = 2\sqrt x + 4 \cr} \)
\( \Leftrightarrow \sqrt x = 4 \Leftrightarrow x = 16(tm)\)
Vậy với \(x=16\) thì \(P=2.\)
