Áp dụng Cô-si:
\(A\le3x+\frac{10-x^2+1}{2}=3x+\frac{11-x^2}{2}\)
\(=\frac{-x^2+6x+11}{2}=\frac{-\left(x^2-6x-11\right)}{2}=\frac{20-\left(x-3\right)^2}{2}\le\frac{20}{2}=10\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}10-x^2=1\\x-3=0\end{matrix}\right.\)\(\Leftrightarrow x=3\)