ĐKXĐ: \(0\le x\le2\)
Đặt \(\sqrt{-x^2+2x}=a\Rightarrow0\le a\le1\)
BPT trở thành: \(-a^2+a-3+m\le0\)
\(\Rightarrow a^2-a+3\ge m\) (1)
Để (1) có nghiệm \(\Rightarrow m\le\max\limits_{\left[0;1\right]}\left(a^2-a+3\right)\)
Đặt \(f\left(a\right)=a^2-a+3\)
\(f\left(0\right)=3\) ; \(f\left(1\right)=3\); \(f\left(\frac{1}{2}\right)=\frac{11}{4}\)
\(\Rightarrow\max\limits_{\left[0;1\right]}f\left(a\right)=3\Rightarrow m\le3\)