Thiếu điều kiện:`x> -3`
`P=(x^2+16)/(x+3)`
`=(x^2-9+25)/(x+3)`
`=x-3+25/(x+3)`
`=x+3+25/(x+3)-6`
Áp dụng bđt cosi ta có:
`x+3+25/(x+3)>=10`
`=>P>=10-6=4`
Dấu "=" xảy ra khi `(x+3)^2=25`
`<=>` \(\left[ \begin{array}{l}x+3=5\\x+3=-5\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2\\x=-8(l)\end{array} \right.\)
Vậy `min_P=4<=>x=2.`