\(N=-\left(x-6\sqrt{x}+9\right)+8=-\left(\sqrt{x}-3\right)^2+8\le8\)
\(N_{max}=8\) khi \(x=9\)
Có: N=6\( \sqrt{x}\)-\(x-1\)
=-(\(-\left(x-6\sqrt{x}+1\right)\)
=\(-\left[\left(\sqrt{x}\right)^2-2.3\sqrt{x}+9-8\right]\)
=\(-\left(\sqrt{x}-3\right)^2+8\)
Có: \(\left(\sqrt{x}-3\right)^2\ge0\forall x\)
=> \(-\left(\sqrt{x}-3\right)^2\le0\forall x\)
=>\(-\left(\sqrt{x}-3\right)^2+8\le8\forall x\)
=>N ≤ 8∀\(x\)
Dấu "=" xảy ra <=> \(\sqrt{x}-3=0\)
<=>\(\sqrt{x}=3\)
<=>\(x=9\)
Vậy MaxN=8 khi \(x=9\)