\(N=6\sqrt{x}-x-1=8-\left(x-6\sqrt{x}+9\right)=8-\left(\sqrt{x}-3\right)^2\le8\)
Dấu "=" xảy ra <=> \(\sqrt{x}-3=0\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\)
Vậy Max(N)=8
\(P=\frac{1}{x-\sqrt{x}+1}=\frac{1}{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}}\le\frac{1}{\frac{3}{4}}=\frac{4}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-\frac{1}{2}=0\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\)
Vậy Max(P)=4/3
\(\sqrt{x-1}\ge0,\forall x\inℝ\Rightarrow\sqrt{3}-\sqrt{x-1}\le\sqrt{3}\)
Dấu "=" xảy ra \(\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy Max (M)=\(\sqrt{3}\)\(\Leftrightarrow x=1\)
