ta có: \(\left(\sqrt{x}+1\right)^2>\sqrt{x}+n\)(đk: x \(\ge0\))
\(\Leftrightarrow x+2\sqrt{x}+1>\sqrt{x}+n\)
\(\Leftrightarrow x+\sqrt{x}+1-n>0\)
\(\Leftrightarrow\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}-n>0\)
vì \(\left(\sqrt{x}+\frac{1}{2}\right)^2\ge0\) nên để \(\left(\sqrt{x}+1\right)^2>\sqrt{x}+n\)thì:
\(\frac{3}{4}-n>0\Leftrightarrow n< \frac{3}{4}\)
vậy n<3/4 thì \(\left(\sqrt{x}+1\right)^2>\sqrt{x}+n\)