ĐKXĐ: \(x\le2\)
Đặt \(\sqrt{2-x}=a>0\) \(\Rightarrow2-x=a^2\Rightarrow x=2-a^2\)
\(\Rightarrow N=2-a^2+a=-a^2+2.a.\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{9}{4}\)
\(\Rightarrow N=-\left(a-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\)
\(\Rightarrow N_{max}=\dfrac{9}{4}\) khi \(a=\dfrac{1}{2}\Leftrightarrow\sqrt{2-x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{7}{4}\)