Giả sử \(P\ge Q\)
\(\Leftrightarrow\sqrt{n}+\sqrt{n+m}\ge\sqrt{n+1}+\sqrt{n+m-1}\)
\(\Leftrightarrow\left(\sqrt{n}+\sqrt{n+m}\right)^2\ge\left(\sqrt{n+1}+\sqrt{n+m-1}\right)^2\)
\(\Leftrightarrow n+2\sqrt{n}\sqrt{n+m}+n+m\ge n+1+2\sqrt{n+1}\sqrt{n+m-1}+n+m-1\)
\(\Leftrightarrow\sqrt{n}\sqrt{n+m}\ge\sqrt{n+1}\sqrt{n+m-1}\)
\(\Leftrightarrow n\left(n+m\right)\ge\left(n+1\right)\left(n+m-1\right)\)
\(\Leftrightarrow n^2+nm\ge n^2+nm-n+n+m-1\)
\(\Leftrightarrow m\le1\)
Vậy \(\left\{{}\begin{matrix}P=Q\left(m=1\right)\\P< Q\left(m\ge2\right)\end{matrix}\right.\)