ta có
\(\sqrt{\left(x-5\right).1}\le\frac{x-5+1}{2}=\frac{x-4}{2}\)
\(\sqrt{\left(7-x\right).1}\le\frac{7-x+1}{2}=\frac{-x+8}{2}\)
\(\Rightarrow P\ge\frac{x-4}{2}+\frac{8-x}{2}=2\)
Dấu = xảy ra <=> \(\hept{\begin{cases}x-5=1\\7-x=1\end{cases}\Leftrightarrow x=6}\)
vậy min P=2 khi x=6