Đáp án:

\(\begin{array}{l} a)\,\,\,Max\,\,M = \sqrt 3 \,\,\,khi\,\,\,x = 1.\\ b)\,\,Max\,\,N = 8\,\,\,khi\,\,\,x = 9.\\ c)\,\,Ma\,xP = - \frac{4}{5}\,\,\,khi\,\,\,x = \frac{1}{4}. \end{array}\)

Giải thích các bước giải: \(\begin{array}{l} a)\,\,M = \sqrt 3 - \sqrt {x - 1} \\ DK:\,\,\,x - 1 \ge 0 \Leftrightarrow x \ge 1\\ Do\,\,\sqrt {x - 1} \ge 0\,\,\forall x \ge 1\\ \Rightarrow - \sqrt {x - 1} \le 0\,\,\forall x \ge 1\\ \Rightarrow \sqrt 3 - \sqrt {x - 1} \le \sqrt 3 \,\,\forall x \ge 1\\ Dau\,\, = \,\,\,xay\,\,ra \Leftrightarrow x - 1 = 0 \Leftrightarrow x = 1\,\,\,\left( {tm} \right).\\ Vay\,\,Max\,\,M = \sqrt 3 \,\,\,khi\,\,\,x = 1.\\ b)\,\,N = 6\sqrt x - x - 1 = - \left( {x - 6\sqrt x + 9} \right) + 9 - 1\,\,\,\,\left( {x \ge 0} \right)\\ = - {\left( {\sqrt x - 3} \right)^2} + 8\\ Vi\,\,\,{\left( {\sqrt x - 3} \right)^2} \ge 0\,\,\forall x \ge 0\\ \Rightarrow - {\left( {\sqrt x - 3} \right)^2} \le 0\,\,\forall x \ge 0\\ \Rightarrow - {\left( {\sqrt x - 3} \right)^2} + 8 \le 8\,\,\forall x \ge 0\\ Dau\,\, = \,\,xay\,\,ra \Leftrightarrow \sqrt x - 3 = 0 \Leftrightarrow x = 9\,\,\,\left( {tm} \right)\\ Vay\,\,\,Max\,\,N = 8\,\,\,khi\,\,\,x = 9.\\ c)\,\,P = \frac{1}{{x - \sqrt x - 1}}\,\,\,\,\left( {x \ge 0} \right)\\ Ta\,\,co:\,\,x - \sqrt x - 1 = x - 2.\frac{1}{2}\sqrt x + \frac{1}{4} - \frac{1}{4} - 1\\ = {\left( {\sqrt x - \frac{1}{2}} \right)^2} - \frac{5}{4}\\ Vi\,\,{\left( {\sqrt x - \frac{1}{2}} \right)^2} \ge 0\,\,\,\forall x \ge 0 \Rightarrow {\left( {\sqrt x - \frac{1}{2}} \right)^2} - \frac{5}{4} \ge - \frac{5}{4}\,\,\,\forall x \ge 0\\ \Rightarrow \frac{1}{{{{\left( {\sqrt x - \frac{1}{2}} \right)}^2} - \frac{5}{4}}} \le \frac{1}{{ - \frac{5}{4}}} = - \frac{4}{5}\\ Dau\,\, = \,\,xay\,\,ra \Leftrightarrow \sqrt x - \frac{1}{2} = 0 \Leftrightarrow x = \frac{1}{4}.\\ Vay\,\,Ma\,xP = - \frac{4}{5}\,\,\,khi\,\,\,x = \frac{1}{4}. \end{array}\)